Previous year questions (2013–2025) grouped by marks — 2, 3, 4 & 5 mark questions with answers. 176 questions total.
Chapter-wise Question Distribution
Chapter
2M
3M
4M
5M
Total
Solutions
25
26
1
8
60
Electrochemistry
17
16
3
14
50
Chemical Kinetics
21
19
1
6
47
Solid State
5
5
–
1
11
Surface Chemistry
2
6
–
–
8
Total
70
72
5
29
176
3 Mark Questions ★
Rate Law / Order / MolecularityAsked 17 times
201420162017201820192020202320242025
2014 · 3 marks
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume: SO2Cl2(g) → SO2(g) + Cl2(g) Experiment 1: Time = 0, Total pressure = 0.4 atm Experiment 2: Time = 100s, Total pressure = 0.7 atm Calculate the rate constant. (Given: log 4 = 0.6021, log 2 = 0.3010)
AnswerAt t = 0: Pi = 0.4 atm (only SO2Cl2)
At t = 100s: Total pressure = 0.7 atm
PSO2Cl2 at time t = 2Pi − Pt = 2(0.4) − 0.7 = 0.1 atm
k = (2.303/t) × log(Pi/(2Pi − Pt))
k = (2.303/100) × log(0.4/0.1)
k = 2.303 × 10−2 × log 4
k = 2.303 × 10−2 × 0.6021
k = 1.3866 × 10−2 s−1
2016 · 3 marks
The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.2 - (1.0 x 104)/T K; Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 J K-1 mol-1)
AnswerComparing log k = log A - Ea/(2.303RT) with log k = 14.2 - 104/T: Ea/(2.303R) = 1.0 x 104. Ea = 1.0 x 104 x 2.303 x 8.314 = 191471.4 J/mol = 191.47 kJ/mol. For half-life: t1/2 = 0.693/k, k = 0.693/200 = 3.47 x 10-3 min-1.
2016 · 3 marks
For the first order thermal decomposition reaction: C2H5Cl(g) -> C2H4(g) + HCl(g). Time/sec: 0, 300. Total pressure/atm: 0.30, 0.50. Calculate the rate constant. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
AnswerP0 = 0.30, Pt = 0.50, t = 300 s. Pressure of C2H5Cl at time t = 2P0 - Pt = 0.60 - 0.50 = 0.10 atm. k = (2.303/t) log(P0/(2P0 - Pt)) = (2.303/300) log(0.30/0.10) = (2.303/300) x 0.4771 = 3.66 x 10-3 s-1.
Adsorption / Colloids / EmulsionsAsked 12 times
201420162017201820192020
2014 · 3 marks
(a) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(b) Write an important characteristic of lyophilic sols.
(c) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid.
Answer(a) x/m = kP1/n (where x = mass of adsorbate, m = mass of adsorbent, P = pressure, k and n are constants, 1/n is between 0 and 1)
(b) Lyophilic sols are reversible in nature. They are self-stabilizing and do not need stabilizing agents.
(c) Associated colloid: Soap (sodium stearate) forms micelles. Multimolecular colloid: Gold sol (Au atoms aggregate).
2016 · 3 marks
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl3, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multi-molecular colloids?
Answer(i) Adsorption: Surface phenomenon where molecules accumulate on the surface. Absorption: Bulk phenomenon where molecules are uniformly distributed throughout the body.
(ii) AlCl3 is more effective because Al3+ has higher charge (+3) than Mg2+ (+2). Higher the charge of coagulating ion, greater its coagulating power (Hardy-Schulze rule).
(iii) Sulphur sol forms multi-molecular colloids.
2016 · 3 marks
Define the following terms:
(i) Lyophilic colloid
(ii) Zeta potential
(iii) Associated colloids.
Answer(i) Lyophilic colloid: Colloids in which the dispersed phase has strong affinity for the dispersion medium. They are reversible and self-stabilizing. E.g., starch, gum in water.
(ii) Zeta potential: The potential difference between the fixed charged layer and the diffuse layer of the electrical double layer surrounding a colloidal particle (electrokinetic potential).
(iii) Associated colloids: Substances which behave as normal electrolytes at low concentration but above CMC form micelles and behave as colloids. E.g., soap, detergents.
Solid State / Crystal Structure / DefectsAsked 7 times
20152016201720182019
2015 An element with molar mass 27 g mol 1 forms a cubic unit cell with edge length 4.05 10 8 cm. If its density is 2.7 g cm 3 , what is the nature of the cubic unit cell?
2016 · 3 marks
An element crystallizes in a fcc lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 x 1024 atoms.
AnswerFor FCC, Z = 4. Molar mass: M = 300 x 6.022 x 1023 / (2 x 1024) = 90.33 g/mol. a = 250 pm = 2.5 x 10-8 cm. Density = ZM/(a3 x NA) = 4 x 90.33/((2.5 x 10-8)3 x 6.022 x 1023) = 38.4 g/cm3.
2016 · 3 marks
An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm-3. How many atoms are present in 300 g of the element?
AnswerFor BCC, Z = 2. d = ZM/(a3 x NA). 7.5 = 2M/((500 x 10-10)3 x 6.022 x 1023). M = 7.5 x 75.275/2 = 282.28 g/mol. Number of atoms = (300/282.28) x 6.022 x 1023 = 6.4 x 1023 atoms.
Other (Electrochemistry)Asked 6 times
2014201620172025
2014 (a) Calculate ΔG 0 for the reaction: Mg(s) + Cu 2+ (aq) Mg 2+ (aq) + Cu(s) Given: E 0 cell = 2.71 V, 1 F = 96500 C mol 1 (b) Name the type of cell that was used in Apollo spa...
2016 · 3 marks
Calculate e.m.f. of the following cell at 298 K: 2Cr(s) + 3Fe2+(0.1M) -> 2Cr3+(0.01M) + 3Fe(s). Given: E°(Cr3+|Cr) = -0.74 V, E°(Fe2+|Fe) = -0.44 V.
(a) The cell in which the following reaction occurs: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(s) has E°cell = 0.236 V at 298 K. Calculate standard Gibb's energy of the cell reaction. (Given: 1 F = 96,500 C mol-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1 F = 96,500 C mol-1)
Answer(a) ΔG° = -nFE°cell = -2 x 96500 x 0.236 = -45548 J/mol = -45.55 kJ/mol
(b) Q = It = 0.5 x 2 x 60 x 60 = 3600 C. Number of electrons = (6.023 x 1023 x 3600) / 96500 = 2.25 x 1022 electrons.
Other (Solutions)Asked 6 times
20152016201920202023
2015 · 3 marks
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). (Given: Molar mass of benzoic acid = 122 g mol−1, Kf for benzene = 4.9 K kg mol−1)
1.62 = i × 4.9 × (3.9 × 1000)/(122 × 49) 1.62 = i × 4.9 × 3.9 × 1000/(5978) 1.62 = i × 3.196 i = 0.506
Since i < 1, the solute gets associated. Benzoic acid dimerizes in benzene due to intermolecular hydrogen bonding.
2016 · 3 marks
Calculate the boiling point of solution when 4 g of MgSO4 (M = 120 g/mol) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization. (Kb for water = 0.52 K kg/mol)
AnswerMgSO4 -> Mg2+ + SO42-, i = 2. delta-Tb = i x Kb x m = 2 x 0.52 x (4 x 1000)/(120 x 100) = 2 x 0.52 x 0.333 = 0.346 K. Boiling point = 373.15 + 0.346 = 373.496 K.
2019 · 3 marks
A solution containing 1.9 g per 100 mL of KCl (M = 74.5) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60). Calculate the degree of dissociation of KCl solution.
Answerpi(urea) = pi(KCl). (3/60) = i x (1.9/74.5). 0.05 = i x 0.0255. i = 1.96. For KCl (n=2): alpha = (i-1)/(n-1) = (1.96-1)/(2-1) = 0.96 or 96%.
Nernst Equation / EMF CalculationAsked 4 times
201320152024
2013 Calculate the emf of the following cell at 298 K: Fe(s) | Fe 2+ (0.001 M) || H + (1M) | H 2 (g) (1 bar), Pt(s) (Given E cell = +0.44 V)
2015 Calculate emf of the following cell at 25 C: Fe | Fe 2+ (0.001 M) || H + (0.01 M) | H 2 (g)(1 bar) | Pt(s) E (Fe 2+ |Fe) = 0.44 V, E (H + |H 2 ) = 0.00 V
2024 Calculate the emf of the cell: Ni(s) + 2Ag+ (0.01 M) -> Ni2+ (0.1 M) + 2Ag(s) Given that E°cell = 1.05 V, log 10 = 1
Colligative Properties (Boiling/Freezing Point, Osmotic Pressure)Asked 3 times
201320182025
2013 · 3 marks
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10−2 g of K2SO4 in 2L of water at 25°C, assuming that it is completely dissociated. (R = 0.0821 L atm K−1 mol−1, Molar mass of K2SO4 = 174 g mol−1)
AnswerK2SO4 → 2K+ + SO42− Number of ions produced (i) = 3 π = iCRT = i × (n/V) × R × T π = 3 × (0.025/174) × 0.0821 × 298 / 2
= 3 × 0.025 × 0.0821 × 0.5 × 298 / 174
= 5.27 × 10−3 atm
2018 Give reasons for the following: (a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers. (b) A...
2025 · 3 marks
A solution containing 15 g urea (molar mass = 60 g mol-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol-1) in water. Calculate the mass of glucose present in one litre of its solution.
AnswerFor isotonic solutions, pi_urea = pi_glucose. Since pi = CRT and RT is common, C_urea = C_glucose. Moles of urea = 15/60 = 0.25 mol for 1 litre. n_glucose = 0.25 mol. Mass of glucose = 0.25 x 180 = 45 g. Hence, the mass of glucose present in one litre of its solution is 45 g.
Carbohydrates (Glucose, Sucrose, Starch)Asked 3 times
201720192025
2017 · 3 marks
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1)
(Molar mass of glucose = 180 g mol-1)
AnswerΔTf = (273.15 - 269.15) = 4 K m (sucrose) = (10/342) x (1000/90) = 0.3244 mol kg-1 Kf = ΔTf/m = 4/0.3244 = 12.33 K kg mol-1 m (glucose) = (10/180) x (1000/90) = 0.6166 mol kg-1
ΔTf = 12.33 x 0.6166 = 7.60 K
Tf = 273.15 - 7.60 = 265.55 K
2019 · 3 marks
A 4% solution w/w of sucrose (M = 342 g mol-1) in water has a freezing point of 271.15K. Calculate the freezing point of 5% glucose (M = 180 g mol-1) in water. (Given: Freezing point of pure water = 273.15 K)
AnswerFor sucrose: Delta_Tf = 273.15 - 271.15 = 2.00 K. Molality = (4/342)/(96/1000) = 0.1218 m. Kf = Delta_Tf/m = 2.00/0.1218 = 16.42 K kg mol-1. For glucose: Molality = (5/180)/(95/1000) = 0.2923 m. Delta_Tf = 16.42 x 0.2923 = 4.801 K. Freezing point = 273.15 - 4.801 = 268.35 K.
2025 · 3 marks
A solution of glucose (molar mass = 180 g mol-1) in water has a boiling point of 100.20 C. Calculate the freezing point of the same solution. Molal constants for water Kf and Kb are 1.86 K kg mol-1 and 0.512 K kg mol-1 respectively.
AnswerTb of glucose solution = 100.20 C, delta Tb = 100.20 - 100 = 0.20 C. delta Tb = Kb x m, m = 0.20/0.512 = 0.390 mol/kg. delta Tf = Kf x m = 1.86 x 0.390 = 0.725 C. Freezing point of solution = 0 - 0.725 = -0.725 C.
5 Mark Questions ★★★
Rate Law / Order / MolecularityAsked 6 times
201320152020
2013 · 5 marks
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 of this reaction. (Given log 1.428 = 0.1548)
Answer(a)
(i) Rate = k[A]2[B]
(ii) If [A] increases 3 times: Rate2/Rate1 = (3a)2 × b / (a2 × b) = 9. Rate becomes 9 times.
(iii) If [A] and [B] are doubled: Rate2/Rate1 = (2a)2 × (2b) / (a2 × b) = 8. Rate becomes 8 times.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
(b) Rate constant 'k' of a reaction varies with temperature 'T' according to the equation:
log k = log A − Ea/(2.303R) × (1/T)
Where Ea is the activation energy. When a graph is plotted for log k Vs. 1/T, a straight line with a slope of −4250 is obtained. Calculate Ea for the reaction. (R = 8.314 JK−1 mol−1)
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained: t/s: 0, 30, 60 [CH3COOCH3]/mol L−1: 0.60, 0.30, 0.15
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
Answer(i) k = (2.303/t) log([A0]/[A])
At t = 30: k = (2.303/30) log(0.60/0.30) = (2.303/30) × 0.3010 = 0.023 s−1
At t = 60: k = (2.303/60) log(0.60/0.15) = (2.303/60) × 0.6021 = 0.023 s−1
Since k is constant in both readings, it is a pseudo-first order reaction.
(ii) Average rate = −Δ[R]/Δt = −(0.15 − 0.30)/(60 − 30) = 0.15/30 = 0.005 mol L−1 s−1
Colligative Properties (Boiling/Freezing Point, Osmotic Pressure)Asked 6 times
20142016201720232024
2014 (a) Define the following terms: (i) Molarity (ii) Molal elevation constant (K b ) (b) A solution containing 15 g of urea (molar mass = 60 g mol 1 ) per litre of solution in water h...
2016 · 5 marks
(a) Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g/mol) was dissolved in 50 g of water, assuming complete ionization. (Kf for water = 1.86 K kg/mol). (b)
(i) Out of 1M glucose and 2M glucose, which has higher boiling point and why?
(ii) What happens when external pressure exceeds osmotic pressure of solution?
OR
(a) When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx). (Kf for CS2 = 3.83 K kg/mol, Atomic mass of S = 32 g/mol).
(b) Blood cells are isotonic with 0.9% NaCl. What happens in:
(i) 1.2% NaCl?
(ii) 0.4% NaCl?
Answer(a) MgCl2 -> Mg2+ + 2Cl-, i = 3. m = (1.9 x 1000)/(95 x 50) = 0.4 mol/kg. delta-Tf = 3 x 1.86 x 0.4 = 2.232 K. Tf = 273.15 - 2.232 = 270.918 K. (b)
(i) 2M glucose has higher boiling point due to greater elevation.
(ii) Reverse osmosis takes place.
OR
(a) MB = (3.83 x 2.56 x 1000)/(0.383 x 100) = 256 g/mol. n = 256/32 = 8. Formula: S8. (b)
(i) Cells shrink (crenation) - hypertonic.
(ii) Cells swell (haemolysis) - hypotonic.
2017 · 5 marks
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass
OR
(a) 30 g of urea (M = 60 g mol-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
(b) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) | Cu2+(0.01 M) | Cu(s) [Give E0cell = ± 2.71 V, F = 96500 C mol−1]
Answer(a) Faraday's first law: The mass of substance deposited at any electrode is directly proportional to the amount of charge passed (m = Z × Q). Cu2+ + 2e− → Cu. Since 2 mol of electrons are involved, 2F charge is required.
(b) Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Ecell = E°cell − (2.303RT/nF) log([Mg2+]/[Cu2+])
= 2.71 − (0.0591/2) × log(0.1/0.01)
= 2.71 − (0.0591/2) × log 10
= 2.71 − 0.02955 = 2.68 V
2019 · 5 marks
(a) Conductivity of 0.001 mol/L acetic acid is 4.95x10-5 S cm-1. Calculate dissociation constant if Lambda_m0 for acetic acid is 390.5 S cm2 mol-1.
(b) Strong: Lambda_m decreases slightly with concentration. Weak: Lambda_m increases sharply at low concentration. Lambda_m0 for weak electrolyte obtained by Kohlrausch's law: Lambda_m0 = v+*lambda_+0 + v-*lambda_-0.
Electrochemical / Galvanic CellAsked 4 times
201620182019
2016 · 5 marks
(a) Calculate E°cell for the following reaction at 298K: 2Al(s) + 3Cu2+(0.01M) -> 2Al3+(0.01M) + 3Cu(s). Given: E°cell = 1.98 V.
(b) Using E° values of A and B, predict which is better for coating iron [E°(Fe2+/Fe) = -0.44V] to prevent corrosion. Given: E°(A2+/A) = -2.37 V, E°(B2+/B) = -0.14 V.
OR
(a) The conductivity of 0.001 mol/L solution of CH3COOH is 3.905 x 10-5 S/cm. Calculate molar conductivity and degree of dissociation. Given lambda0(H+) = 349.6, lambda0(CH3COO-) = 40.9 S cm2/mol.
(b) Define electrochemical cell. What happens if external potential exceeds E°cell?
(b) Metal A (E° = -2.37 V) is better as it is more negative than Fe; it gets oxidized preferentially (sacrificial protection).
OR
(a) Lambda_m = 3.905 x 10-5 x 1000/0.001 = 39.05 S cm2/mol. Lambda0 = 349.6 + 40.9 = 390.5. alpha = 39.05/390.5 = 0.1 (10%).
(b) Electrochemical cell produces electricity from spontaneous reactions. When external potential exceeds E°cell, reaction reverses (becomes electrolytic cell).
2018 · 5 marks
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: Sn(s) | Sn2+(0.004 M) || H+(0.020 M) | H2(g)(1 bar) | Pt(s) (Given: E°Sn2+/Sn = -0.14 V)
(b) Give reasons:
(i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH3COOH decreases on dilution.
OR
(a) For the reaction: 2AgCl(s) + H2(g)(1 atm) → 2Ag(s) + 2H+(0.1 M) + 2Cl-(0.1 M), ΔG° = -43600 J at 25°C. Calculate the e.m.f. of the cell. [log 10-n = -n]
(b) Define fuel cell and write its two advantages.
(i) Due to overpotential/overvoltage of O2, Cl2 is liberated at lower applied voltage.
(ii) Conductivity decreases because number of ions per unit volume decreases on dilution.
OR
(a) ΔG° = -nFE°. -43600 = -2 x 96500 x E°. E° = 0.226 V
E = 0.226 - (0.059/2) log[(0.1)2 x (0.1)2/1] = 0.226 - 0.059/2 x log 10-4 = 0.226 + 0.118 = 0.344 V
(b) Fuel cells convert energy of combustion of fuels directly into electrical energy. Advantages: High efficiency, non-polluting.
2019 · 5 marks
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited. (Molar mass: Fe = 56, Zn = 65.3, 1F = 96500 C mol-1)
(b) In the plot of molar conductivity vs. square root of concentration, following curve obtained for two electrolytes A and B:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of Lambda_m to concentration approaching zero for electrolytes A and B?
Answer(a) m = zIt/F. 2.8 = (56 x 2 x t)/(2 x 96500). t = 4825 s. Mass of Zn = 2.8 x (65.3/2)/(56/2) = 3.265 g. (b)
(i) A is strong electrolyte (linear), B is weak electrolyte (curved).
(ii) For strong electrolyte A, Lambda_m0 can be obtained by extrapolation. For weak electrolyte B, Lambda_m0 cannot be obtained by extrapolation as curve rises steeply.
2014 · 2 marks
Write two differences between 'order of reaction' and 'molecularity of reaction'.
Answer(i) Molecularity is the total number of reactant molecules in an elementary reaction. Order is the sum of powers of concentration terms in the rate law.
(ii) Molecularity is a theoretical concept and can be calculated by simple adding the moles. Order can only be obtained analytically and not through computation.
2014 · 2 marks
Explain the following terms:
(i) Rate constant (k)
(ii) Half life period of reaction (t1/2).
Answer(i) Rate constant (k): The proportionality constant that links the pace of the reaction to the concentrations of reactants is known as the specific rate constant. Temperature affects the rate constant's value.
(ii) Half-life period (t1/2): The amount of time needed for a reactant to achieve half of its starting concentration or pressure is known as the half-life of a reaction. A first-order reaction's half-life is stable over time and concentration-independent. t1/2 = 0.693/k.
2014 · 2 marks
For a chemical reaction R → P, the variation in the concentration (R) vs. time (t) plot is given as a straight line with negative slope.
(i) Predict the order of the reaction.
(ii) What is the slope of the curve?
Answer(i) The change in concentration (R) vs. time (t) as a straight line shows this is a zero order reaction, for which [R] = [R]0 − kt.
(ii) The slope of the curve representing the fluctuation in [R] vs t plot is opposite the reaction's rate constant, i.e., slope = −k.
2016 · 2 marks
For a reaction: 2NH3(g) --(Pt catalyst) → N2(g) + 3H2(g); Rate = k;
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer(i) Order = Zero (since Rate = k, independent of concentration). Molecularity = bimolecular.
(ii) Unit of k = mol L-1 s-1.
2016 · 2 marks
For a reaction: H2 + Cl2 --(hv) → 2HCl; Rate = k;
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer(i) Order = Zero (Rate = k). Molecularity = 2 (bimolecular).
(ii) Unit of k = mol L-1 s-1.
2018 · 2 marks
For the reaction 2N2O5(g) → 4NO2(g) + O2(g), the rate of formation of NO2 is 2.8 x 10-3 M s-1. Calculate the rate of disappearance of N2O5(g).
For a reaction, 2H2O2 --(I-/alkaline medium) → 2H2O + O2, the proposed mechanism is: (1) H2O2 + I- -> H2O + IO- (slow), (2) H2O2 + IO- -> H2O + I- + O2 (fast).
(i) Write rate law of the reaction.
(ii) Write overall order of reaction.
(iii) Out of step (1) and (2), which one is rate determining step?
Answer(i) Rate = k[H2O2][I-].
(ii) Order = 2.
(iii) Step 1 is the rate determining step as it is the slow step.
2019 · 2 marks
Define order of reaction. Predict the order of reaction in the given graphs:
(a) t1/2 vs [R]0 showing straight line through origin.
(b) t1/2 vs [R]0 showing horizontal line.
AnswerOrder of reaction: Sum of powers to which concentration terms are raised in rate law.
(a) First order (t1/2 independent of initial concentration, but per CBSE marking scheme graph interpretation).
(b) Zero order.
2022-II · 2 marks
Answer the following questions (Do any two):
(a) Identify the order of reaction from the following unit for its rate constant: L mol-1 s-1.
(b) The conversion of molecules A to B follow second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B?
(c) Write the expression of integrated rate equation for zero order reaction.
Answer(a) The unit L mol-1 sec-1 for rate constant is the unit of second order reaction.
(b) For reaction A -> B, Rate r = k[A]2. If concentration increased to 3 times: r' = k[3A]2. r/r' = 1/9. Therefore, rate of formation of B increases to nine times.
(c) Integrated rate equation for zero order reaction is: k = [R0 - R] / t, where k = rate constant, R0 = Initial concentration, R = Final concentration, t = time taken.
2023 · 2 marks
What happens to the rate constant k and activation energy Ea as the temperature of a chemical reaction is increased? Justify.
AnswerRate constant increases with the increase in temperature because rate of the reaction increases. The rate of the reaction becomes doubled after every ten degree rise in temperature. The activation energy also increases with increase in temperature because kinetic energy of the molecules increases their colliding frequency will be very high and activation energy increases.
2023 · 2 marks
(a) The conversion of molecule A to B followed second order kinetics. If concentration of A increased to three times, how will it affect the rate of formation of B?
AnswerReaction is A -> B for second order kinetics: (Rate)1 = K[A]2. (Rate)2 = K[3A]2 = 9K[A]2. (Rate)2 = 9 x (Rate)1. So the rate of formation of B will increase by 9 times.
2023 · 2 marks
(b) Define Pseudo first order reaction with an example. (OR)
AnswerThe reactions that have higher order true rate law but are found to behave as first order are known as pseudo first order reactions. Example: Consider the acid hydrolysis of methyl acetate. CH3COOCH3(aq) + H2O(l) -> CH3COOH(aq) + CH3OH(aq)
2025 · 2 marks
(A) The rate constant for a zero order reaction A->P is 0.0030 mol L-1 s-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? OR
(B) The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 x 10-4 mol L-1 s-1?
Answer(A) K = 0.0030 mol L-1 s-1. [A0] = 0.10 M, [A] = 0.075 M. For zero-order: K = ([A0] - [A])/t. 0.0030 = (0.10 - 0.075)/t. t = 0.025/0.0030 = 8.33 sec. OR
(B) For zero-order reaction, rate of reaction = K. 2NH3 -> N2 + 3H2. Rate of production of N2 = 1/3 rate = k/1 but considering stoichiometry: Rate = k = 2.5 x 10-4. Rate of production of N2 = k x (1/2) but rate of reaction = rate of production of N2 = 1/3 rate of H2. Rate of N2 = 3 x 2.5 x 10-4 = 7.5 x 10-4 mol L-1 s-1. Actually: Rate of production of H2 = 3k = 7.5 x 10-4 mol L-1 s-1. Rate of production of N2 = k = 2.5 x 10-4 mol L-1 s-1.
2025 · 2 marks
Define the following terms:
(a) Pseudo first order reaction
(b) Half-life period of reaction (t1/2)
Answer(a) A pseudo-first-order reaction is inherently of a higher order but behaves as if it were first-order under certain conditions. This typically occurs when one reactant is present in such a large excess that its concentration remains effectively constant throughout the reaction. Example: CH3COOCH3 + H2O -> CH3COOH + CH3OH.
(b) The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value.
2025 · 2 marks
A reaction is of second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is
(i) doubled
(ii) reduced to half?
AnswerRate = k[A]2.
(i) If concentration is doubled: A -> 2A, R = k[2A]2 = 4kA2. Rate increases 4 times.
(ii) If concentration is reduced to half: A -> A/2, R = k[A/2]2 = kA2/4. Rate becomes 1/4th.
2025 · 2 marks
What is meant by the Rate law and Rate constant of a reaction. Identify the order of a reaction if the units of its Rate constant are:
(a) s-1
(b) mol-1 Ls-1
AnswerRate law is the expression which expresses the rate of reaction in terms of molar concentration of the reactants with each term raised to some power. Rate = k[A]^a[B]^b. Rate constant may be defined as the rate of the reaction when the molar concentration of each reactant is taken as unity.
(a) s-1: First order reaction (n = 1).
(b) mol-1 Ls-1: Second order reaction (n = 2).
2024 · 2 marks
Show that in case of a first order reaction, the time taken for completion of 99% reaction is twice the time required for 90% completion of the reaction. (log 10 = 1)
AnswerFor 99% completion: t = (2.303/k) log(100/1) = (2.303/k) x log 102 = 2 x (2.303/k)
For 90% completion: t' = (2.303/k) log(100/10) = (2.303/k) x log 10 = (2.303/k)
Comparing: t = 2t'. So the time for 99% completion is twice the time for 90% completion.
2024 · 2 marks
A first order reaction takes 40 min for 75% decomposition. Calculate rate constant. [Given: log 2 = 0.30, log 4 = 0.60]
AnswerFor first-order reaction: k = (2.303/t) log(100/(100-x))
k = (2.303/40) log(100/25)
k = (2.303/40) log 4
k = (2.303/40) x 0.60
k = 0.0345 min-1
2024 · 2 marks
Define the following terms:
(a) Order of a reaction
(b) Activation energy
Answer(a) Order of a reaction: The order of a chemical reaction describes how the rate of the reaction depends on the concentration of reactants. It is determined experimentally and may be an integer, fractional, or zero.
(b) Activation Energy: The minimum amount of energy required to initiate a chemical reaction by breaking the bonds of reactant molecules and forming the activated complex or transition state. It is usually denoted by Ea.
2024 · 2 marks
Define the following terms:
(a) Molecularity of reaction
(b) Complex reaction
Answer(a) Molecularity of a chemical reaction refers to the number of molecules or ions participating as reactants in an elementary reaction. It is an integer value that represents the number of entities that come together to undergo a reaction.
(b) Complex reactions involve multiple elementary steps, each with its own reaction mechanism. These reactions can be broken down into simpler steps. The overall reaction is a combination of these individual steps. OR: The reaction in which the molecularity of the reaction is not equal to order of reaction.
2024 · 2 marks
Define the following terms:
(a) Half life period (t1/2)
(b) Effective collisions
Answer(a) Half-life period of a reaction is the time in which the concentration of a reactant is reduced to half of its initial concentration.
(b) Effective collision: The effective collision is that in which molecules collide with sufficient kinetic energy and proper orientation. It leads to breaking of bonds of reactants and formation of new bonds of products.
2013 · 2 marks
The conductivity of 0.20 M solution of KCl at 298K is 0.025 S cm−1. Calculate its molar conductivity.
AnswerMolar conductivity (Λm) = (κ × 1000) / C
= (0.025 × 1000) / 0.20
= 125 S cm2 mol−1
2014 · 2 marks
State the Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
AnswerKohlrausch's law states that the limiting molar conductivity of an electrolyte can be expressed as the sum of the individual contributions made by its cations and anions: Λ0m = ν+λ0+ + ν−λ0− Conductivity decreases with dilution because there are fewer ions available per unit volume for conduction.
2014 · 2 marks
Define the following terms:
(i) Fuel cell
(ii) Limiting molar conductivity (Λ0m)
Answer(i) Fuel cells are galvanic cells or electrochemical cells that transform the chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
(ii) An electrolyte's molar conductivity is said to be limiting when its concentration gets close to zero.
2015 · 2 marks
(a) Following reactions occur at cathode during electrolysis of aqueous silver chloride solution: Ag+(aq) + e− → Ag(s) E° = +0.80 V H+(aq) + e− → ½H2(g) E° = 0.00 V
On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?
(b) Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration?
Answer(a)
(i) Ag+(aq) + e− → Ag(s) is feasible at cathode because it has higher E° value (+0.80 V) / ΔG° is more negative.
(ii) Limiting molar conductivity (Λ0m) is the molar conductivity of a solution at infinite dilution or when concentration approaches zero.
Conductivity decreases with decrease in concentration because number of ions per unit volume decreases.
2016 · 2 marks
From the given cells: Lead storage cell, Mercury cell, Fuel cell and Dry cell. Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?
Answer(i) Mercury cell
(ii) Fuel cell
(iii) Lead storage cell
(iv) Dry cell.
2017 · 2 marks
Calculate the degree of dissociation (α) of acetic acid if its molar conductivity (Λm) is 39.05 S cm2 mol-1.
Given λ°(H+) = 349.6 S cm2 mol-1 and λ°(CH3COO-) = 40.9 S cm2 mol-1.
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.
AnswerMercury cell is generally used in hearing aids (transistors use dry cell). Anode: Zn(Hg) + 2OH- → ZnO(s) + H2O + 2e-. Cathode: HgO + H2O + 2e- → Hg(l) + 2OH-.
2022-II · 2 marks
(a) Why on dilution the molar conductivity (Lambda_m) of CH3COOH increases very fast, while that of CH3COONa increases gradually?
(b) What happens if external potential applied becomes greater than E° of electrochemical cell?
Answer(a)
(i) On dilution Lambda_m of CH3COOH increases very fast because it is a weak electrolyte and the number of ions also increases due to increase in the degree of dissociation (less than 1). However, in CH3COONa which is a strong electrolyte, the number of ions remains same but the intermolecular forces of attraction decreases and the degree of dissociation is equal to 1. Thus, Lambda_m increases gradually in case of CH3COONa.
(b)
(ii) If the external potential applied becomes greater than E° cell of electrochemical cell, then the reaction gets reversed and the electrochemical cell acts as an electrolytic cell and vice-versa.
2023 · 2 marks
(a)
(i) What should be the signs (positive/negative) for E°cell and Delta G° for a spontaneous redox reaction occurring under standard conditions?
(ii) State Faraday's first law of electrolysis.
Answer(a)
(i) For a spontaneous reaction: Delta G° = -ve, E°cell = +ve.
(ii) Faraday's first law of electrolysis: The amount of chemical reaction which occur at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution).
2023 · 2 marks
19 (OR): Calculate the emf of the following cell at 298 K: Fe(s) | Fe2+ (0.01M) || H+(1M) | H2(g) (1 bar), Pt(s)
Given E°cell = 0.44 V.
AnswerAccording to the equation: Fe(s) + 2H+(aq) -> Fe2+(aq) + H2(g)
By applying Nernst Equation:
Ecell = E°cell - (0.0591/n) log([Fe2+]/[H+]2)
= 0.44 - (0.0591/2) log(0.001/(1))
= 0.44 - 0.0295 log 10-3
= 0.44 - 0.0295(-3 log 10)
= 0.44 + 0.089
Ecell = 0.53 V
2023 · 2 marks
(a) Give reasons:
(i) Mercury cell delivers a constant potential during its life time.
(ii) In the experimental determination of electrolytic conductance, Direct Current (DC) is not used.
Answer(a)
(i) As the overall reaction does not involve any ion in the solution whose concentration changes during its life period.
(ii) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell. So, AC current is used for this measurement to prevent its electrolysis.
2023 · 2 marks
(b) Define fuel cell with an example. What advantages do the fuel cells have over primary and secondary batteries? (OR)
AnswerFuel cell: A fuel cell is an electrochemical cell that generates electricity/electrical energy from fuel via an electrochemical reaction. It offers high efficiency and zero emissions. e.g. The polymer electrolyte fuel cells etc. Advantages: Good reliability - quality of power does not degrade over time. Environmentally beneficial - greatly reduces CO2 & harmful pollutant emission.
2024 · 2 marks
State:
(a) Kohlrausch law of independent migration of ions.
(b) Faraday's first law of electrolysis.
Answer(a) Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductance of the anions and cations.
(b) Faraday's first law of electrolysis states, during electrolysis, the amount of substance produced at any electrode as a result of chemical reaction is proportional to the quantity of electricity passed through the electrolyte.
2024 · 2 marks
Resistance of a conductivity cell filled with 0.2 mol L-1 KCl solution is 200 ohm. If the resistance of the same cell when filled with 0.05 mol L-1 KCl solution is 620, calculate the conductivity and molar conductivity of 0.05 mol L-1 KCl solution. The conductivity of 0.2 mol L-1 KCl solution is 0.0248 S cm-1.
AnswerCell constant = conductivity x resistance = 0.0248 x 200 = 4.96 m-1 = 0.0496 cm-1
For 0.05 mol L-1 KCl: Conductivity = cell constant/resistance = 0.0496/620 = 8 x 10-5 ohm-1 cm-1
Molar conductivity = conductivity/concentration = (8 x 10-5)/(0.05 x 10-3) = 160 x 10-2 S cm2 mol-1
2024 · 2 marks
Define the following terms:
(a) Faraday's second law of electrolysis
(b) Corrosion
Answer(a) Faraday's second law of electrolysis: When the same quantity of electricity is passed through the solutions of different electrolytes connected in series, the masses of substances liberated (or deposited) at the electrodes are directly proportional to their equivalent masses.
(b) Corrosion: It is a process of slow deterioration of metals which occurs when most or all of the atoms on the same metal surface are oxidized, damaging the entire surface.
2024 · 2 marks
Define the following terms:
(a) Limiting molar conductivity (0m)
(b) Fuel cell
Answer(a) Limiting molar conductivity: It is defined as the measure of the ability of an electrolyte to conduct electricity at infinite dilution, where the concentration approaches zero. It is denoted by the symbol 0m.
(b) Fuel cell: A fuel cell is an electrochemical device that converts chemical energy directly into electrical energy by combining a fuel (usually hydrogen) and an oxidizing agent (usually oxygen or air) within an electrochemical cell.
2014 · 2 marks
An element with density 11.2 g cm−3 forms f.c.c. lattice with edge length 4 × 10−8 cm. Calculate the atomic mass of the element. (Given NA = 6.022 × 1023 mol−1)
AnswerFor FCC, Z = 4 d = ZM/(a3 × NA) 11.2 = 4 × M / ((4 × 10−8)3 × 6.022 × 1023) M = 11.2 × (64 × 10−24) × 6.022 × 1023 / 4 M = 107.9 g/mol
2014 · 2 marks
Examine the given defective crystal: A+ B− A+ B− A+ B− O B− A+ B− A+ B− A+ O A+ B− A+ B− A+ B− Answer the following questions:
(i) What type of stoichiometric defect is shown by this crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect?
Answer(i) Schottky defect is shown by this crystal (both cation and anion vacancies are present).
(ii) The density of the crystal decreases because of the vacancies created.
(iii) Ionic substances with similar sizes of cations and anions show Schottky defect. E.g., NaCl, KCl, CsCl, AgBr.
2014 · 2 marks
An element with density 2.8 g cm−3 forms a f.c.c. unit cell with edge length 4 × 10−8 cm. Calculate the molar mass of the element. (Given NA = 6.022 × 1023 mol−1)
AnswerFor FCC, Z = 4 d = ZM/(a3NA) M = d × a3 × NA / Z
= 2.8 × (4 × 10−8)3 × 6.022 × 1023 / 4
= 2.8 × 64 × 10−24 × 6.022 × 1023 / 4
= 26.98 g/mol ≈ 27 g/mol
2014 · 2 marks
(i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
(ii) What type of stoichiometric defect is shown by NaCl?
Answer(i) Metal excess defect (F-centre defect). When LiCl is heated in Li vapour, Li atoms deposit on the surface and Cl− diffuses to the surface. The released electron occupies anion vacancy (F-centre) giving pink colour.
(ii) NaCl shows Schottky defect.
2017 · 2 marks
Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a face-centred cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol-1)
AnswerMoles of Al = 8.1/27 = 0.3 mol. Number of atoms = 0.3 x 6.023 x 1023 = 1.8069 x 1023. In FCC, 4 atoms per unit cell. Number of unit cells = 1.8069 x 1023/4 = 4.517 x 1022.
2013 · 2 marks
18 g of glucose, C6H12O6 (Molar Mass = 180 g mol−1) is dissolved in 1 kg of water in a sauce pan. At what temperature will this solution boil? (Kb for water = 0.52 K kg mol−1, boiling point of pure water = 373.15 K)
AnswerΔT = (Kb × WB × 1000) / (mB × WA)
ΔT = (0.52 × 18 × 1000) / (180 × 1000) = 0.052 K
T = 373.15 + 0.052 = 373.202 K
2014 · 2 marks
Calculate the mass of compound (molar mass = 256 g mol−1) to be dissolved in 75 g of benzene to lower its freezing point by 0.48 K (Kf = 5.12 K kg mol−1).
Define an ideal solution and write one of its characteristics.
AnswerAn ideal solution is one that perfectly complies with Raoult's law across the whole concentration range. Such mixtures are created by combining two substances that have almost identical particle sizes, configurations, and intermolecular forces.
(i) Ideal solutions must adhere to Raoult's law.
(ii) There should be no enthalpy of mixing (ΔHmix = 0).
2014 · 2 marks
State Raoult's law for the solution containing volatile components. What is the similarity between Raoult's law and Henry's law?
AnswerAccording to Raoult's Law, the partial pressure of each component in a mixture of volatile liquids equals the product of its vapour pressure at pure state and its mole fraction in the solution. Similarity: Both state that the partial pressure of the volatile component is precisely proportional to its mole fraction. In Raoult's law it is a liquid in a liquid, and in Henry's law it is a gas in a liquid.
2015 · 2 marks
What is meant by positive deviations from Raoult's law? Give an example. What is the sign of ΔmixH for positive deviation? OR Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult's law? Give an example.
AnswerWhen vapour pressure of solution is higher than that predicted by Raoult's law, the intermolecular attractive forces between solute-solvent (A-B) molecules are weaker than those between solute-solute and solvent-solvent molecules (A-A or B-B). E.g., ethanol-acetone / ethanol-cyclohexane / CS2-acetone. ΔmixH is positive. OR Azeotropes are binary mixtures having the same composition in the liquid and vapour phase and boil at a constant temperature. Minimum boiling azeotrope is formed by positive deviation from Raoult's law. E.g., ethanol + water.
2016 · 2 marks
(i) Gas
(A) is more soluble in water than Gas
(B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry's constant) and why?
(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes?
Answer(i) Gas B will have the higher value of KH as lower the solubility of gas, higher is the value of KH (p = KH x).
(ii) Negative deviation from Raoult's law shows the formation of maximum boiling azeotropes.
2017 · 2 marks
Define the following terms:
(i) Colligative properties
(ii) Molality (m)
Answer(i) Colligative properties are those properties of solutions which depend upon the number of solute particles present in the solution irrespective of their nature and are relative to the total number of particles present in the solution. Examples: elevation of boiling point, depression of freezing point.
(ii) Molality is the number of moles of solute dissolved per kilogram of the solvent. m = (Number of moles of solute / Weight of solvent in grams) x 1000
2017 · 2 marks
Define the following terms:
(i) Ideal solution
(ii) Molarity (M)
Answer(i) An ideal solution is one that obeys Raoult's Law over the entire range of concentration. ΔHmix = 0 and ΔVmix = 0.
(ii) Molarity is the number of moles of solute present in 1000 mL of the solution. M = (wB x 1000) / (MB x V(mL))
2017 · 2 marks
Define the following terms:
(i) Abnormal molar mass
(ii) Van't Hoff factor (i)
Answer(i) Abnormal molar mass: When molar mass calculated from colligative properties is different from theoretically expected molar mass due to association or dissociation of molecules.
(ii) Van't Hoff factor (i): Ratio of observed colligative property to calculated colligative property. i = Normal molar mass / Abnormal molar mass.
2018 · 2 marks
Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol-1) in 250 g of water. (Kf of water = 1.86 K kg mol-1)
AnswerΔTf = Kf x m = Kf x (w2 x 1000)/(M2 x w1) = (1.86 x 60 x 1000)/(180 x 250) = 2.48 K Tf = 273.15 - 2.48 = 270.67 K
2019 · 2 marks
State Raoult's law for a solution containing volatile components. Write two characteristics of the solution which obey Raoult's law at all concentrations.
AnswerRaoult's law: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Characteristics of ideal solution:
(i) Delta_mix H = 0.
(ii) Delta_mix V = 0.
(iii) The intermolecular attractions between solute-solute interactions and solvent-solvent interaction are almost similar to the solute-solvent interaction.
2019 · 2 marks
Write two differences between an ideal solution and a non-ideal solution.
AnswerIdeal Solution: (1) Obeys Raoult's law at all concentrations. (2) Delta_mix H = 0, Delta_mix V = 0. Non-ideal Solution: (1) Does not obey Raoult's law. (2) Delta_mix H != 0, Delta_mix V != 0.
2019 · 2 marks
State Henry's law and write its two applications.
AnswerHenry's law: The partial pressure of gas in vapour phase (p) is proportional to mole fraction of gas (x) in solution: p = KH*x. Applications: (1) CO2 is dissolved under high pressure in soft drinks to increase solubility. (2) At high altitudes, low partial pressure of O2 leads to low blood oxygen concentration. (3) Scuba divers must cope with high pressure dissolved gases.
2019 · 2 marks
Give reasons:
(a) Cooking is faster in pressure cooker than in cooking pan.
(b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water.
Answer(a) Due to increase of pressure, boiling point of water increases so food cooks faster.
(b) RBC loses water in saline water (hypertonic) due to osmosis and shrinks. In distilled water (hypotonic), RBC absorbs water and swells.
2019 · 2 marks
Give reasons:
(a) A decrease in temperature is observed on mixing ethanol and acetone.
(b) Potassium chloride solution freezes at a lower temperature than water.
Answer(a) Ethanol-acetone interaction is weaker than pure components (positive deviation). Mixing requires breaking H-bonds, absorbing energy.
(b) KCl dissociates into ions increasing solute particles, causing greater depression in freezing point (colligative property).
2019 · 2 marks
Give reasons:
(a) An increase in temperature is observed on mixing chloroform and acetone.
(b) Aquatic animals are more comfortable in cold water than in warm water.
Answer(a) Chloroform forms strong H-bonding with acetone (negative deviation), releasing energy.
(b) Higher solubility of O2 in cold water (low KH) provides more dissolved oxygen for aquatic life.
2020 · 2 marks
State Raoult's law for a solution containing volatile components. What is the similarity between Raoult's law and Henry's law?
AnswerRaoult's law for a solution containing volatile components states that the partial pressure of a volatile component present in a solution is directly proportional to the mole fraction of that component at a given temperature. PA = K*xA. Raoult's law and Henry's law are similar as both give equation to find partial pressure of gases. PA = KxA (Raoult's law), PA = KH*xA (Henry's law).
2020 · 2 marks
Define adsorption with an example. What is the role of adsorption in heterogeneous catalysis?
OR Define Brownian movement. What is the cause of Brownian movement in colloidal particles? How is it responsible for the stability of Colloidal sol?
AnswerAdsorption is the accumulation of molecular species at the surface rather than in the bulk of a solid or liquid. In heterogeneous catalysis, the reactants are adsorbed on the surface of the catalyst, increasing the concentration of reactants and lowering the activation energy.
OR Brownian movement is the continuous zig-zag movement of colloidal particles. It is caused by unequal bombardment of the particles of the dispersion medium on the colloidal particles. It is responsible for stability as it does not allow the particles to settle down.
2020 · 2 marks
What happens when
(i) a pressure greater than osmotic pressure is applied on the solution side separated from solvent by a semipermeable membrane?
(ii) acetone is added to pure ethanol?
Answer(i) Reverse osmosis will take place and the level of solution will decrease.
(ii) Acetone reacts with alcohol to form a hemiacetal.
2020 · 2 marks
State Henry's law. Calculate the solubility of CO2 in water at 298K under 760 mm Hg. (KH for CO2 in water at 298 K is 1.25 x 106 mm Hg)
AnswerHenry's law: The mass of a gas dissolved in a given volume of the liquid at a constant temperature depends upon the pressure which is applied. xCO2 = Partial pressure of CO2 / KH = 760 / (1.25 x 106) = 608 x 10-6. Mole fraction represents the solubility of CO2 in water.
2023 · 2 marks
The vapour pressure of pure liquid X and pure liquid Y at 25°C are 120 mm Hg and 160 mm Hg respectively. If equal moles of X and Y are mixed to form an ideal solution, calculate the vapour pressure of the solution.
AnswerX = 120 mm Hg, Y = 160 mm Hg. When equal moles of X and Y are mixed, xA = xB = 0.5. According to Raoult's law: P_Total = P0_A * xA + P0_B * xB = 120 * 0.5 + 160 * 0.5 = 60 + 80 = 140 mmHg.
2025 · 2 marks
(A) Give reasons:
(a) Cooking is faster in pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
OR
(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law? Give an example.
Answer(A)
(a) Pressure inside a pressure cooker is artificially increased. As the pressure inside the cooker rises, the boiling point of the liquid also increases. This higher temperature accelerates the cooking process significantly.
(b) On mixing X and Y, if the volume of the solution decreases, it means that the interaction between the X-X and Y-Y molecules reduces and the intermolecular interactions between X and Y increases. This leads to negative deviation of Raoult's law. The temperature of the solution increases which means delta H = -ve. So, it is an exothermic reaction.
OR
(B) Azeotropes are binary mixtures that have the same composition in the liquid and vapour phases and boil at constant temperatures. Azeotropes showing negative deviation from Raoult's law form maximum boiling azeotropes at a specific composition. Example: Azeotrope formed from nitric acid and water.
2024 · 2 marks
18 g of a non-volatile solute is dissolved in 200 g of H2O freezes at 272.07 K. Calculate the molecular mass of solute. (Kf for water = 1.86 K kg mol-1)
AnswerDelta_Tf = 273.15 - 272.07 = 1.08 K Delta_Tf = Kf x m = Kf x (w_B x 1000)/(M_B x W_A) 1.08 = 1.86 x 18 x 1000/(M_B x 200) M_B = 1.86 x 18 x 1000/(1.08 x 200) = 155.01 g/mol
2024 · 2 marks
Calculate the molar mass of a compound when 6.3 g of it is dissolved in 27 g of chloroform to form a solution that has a boiling point of 68.04°C. The boiling point of pure chloroform is 61.04°C and Kb for chloroform is 3.63°C kg mol-1.
AnswerDelta_Tb = 68.04 - 61.04 = 7.0°C M = Kb x w_B x 1000 / (Delta_Tb x W_A) M = 3.63 x 6.3 x 1000 / (7.0 x 27) = 121 g/mol
2024 · 2 marks
A solution containing 60 g of a non-volatile solute in 250 g of water freezes at 270.67 K. Calculate the molar mass of the solute. (Kf of water = 1.86 K kg mol-1)
AnswerDelta_Tf = 273 - 270.67 = 2.33 K M_B = Kf x W_B x 1000 / (Delta_Tf x W_A) M_B = 1.86 x 60 x 1000 / (2.33 x 250) = 191.58 g/mol
2013 · 2 marks
Write the dispersed phase and dispersion medium of the following colloidal systems:
(i) Smoke
(ii) Milk
Answer(i) Smoke: Dispersed phase – Solid (carbon particles), Dispersion medium – Gas (air)
(ii) Milk: Dispersed phase – Liquid (fat), Dispersion medium – Liquid (water)
2013 · 2 marks
Write the differences between physisorption and chemisorption with respect to the following:
(i) Specificity
(ii) Temperature dependence
(iii) Reversibility
(iv) Enthalpy change
Answer(i) Physisorption is not specific; Chemisorption is highly specific.
(ii) Physisorption decreases with increase in temperature; Chemisorption first increases then decreases.
(iii) Physisorption is reversible; Chemisorption is irreversible.
(iv) Physisorption has low enthalpy of adsorption (20-40 kJ/mol); Chemisorption has high enthalpy (80-240 kJ/mol).
2014 · 3 marks
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume: SO2Cl2(g) → SO2(g) + Cl2(g) Experiment 1: Time = 0, Total pressure = 0.4 atm Experiment 2: Time = 100s, Total pressure = 0.7 atm Calculate the rate constant. (Given: log 4 = 0.6021, log 2 = 0.3010)
AnswerAt t = 0: Pi = 0.4 atm (only SO2Cl2)
At t = 100s: Total pressure = 0.7 atm
PSO2Cl2 at time t = 2Pi − Pt = 2(0.4) − 0.7 = 0.1 atm
k = (2.303/t) × log(Pi/(2Pi − Pt))
k = (2.303/100) × log(0.4/0.1)
k = 2.303 × 10−2 × log 4
k = 2.303 × 10−2 × 0.6021
k = 1.3866 × 10−2 s−1
2014 · 3 marks
The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume: SO2Cl2(g) → SO2(g) + Cl2(g) Experiment 1: Time = 0, Total pressure = 0.4 atm Experiment 2: Time = 100s, Total pressure = 0.7 atm Calculate the rate constant. (Given: log 4 = 0.6021, log 2 = 0.3010)
The rate constant for the first order decomposition of H2O2 is given by the following equation: log k = 14.2 - (1.0 x 104)/T K; Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given: R = 8.314 J K-1 mol-1)
AnswerComparing log k = log A - Ea/(2.303RT) with log k = 14.2 - 104/T: Ea/(2.303R) = 1.0 x 104. Ea = 1.0 x 104 x 2.303 x 8.314 = 191471.4 J/mol = 191.47 kJ/mol. For half-life: t1/2 = 0.693/k, k = 0.693/200 = 3.47 x 10-3 min-1.
2016 · 3 marks
For the first order thermal decomposition reaction: C2H5Cl(g) -> C2H4(g) + HCl(g). Time/sec: 0, 300. Total pressure/atm: 0.30, 0.50. Calculate the rate constant. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
AnswerP0 = 0.30, Pt = 0.50, t = 300 s. Pressure of C2H5Cl at time t = 2P0 - Pt = 0.60 - 0.50 = 0.10 atm. k = (2.303/t) log(P0/(2P0 - Pt)) = (2.303/300) log(0.30/0.10) = (2.303/300) x 0.4771 = 3.66 x 10-3 s-1.
2017 · 3 marks
Following data are obtained for the reaction: N2O5 → 2NO2 + 1/2 O2 t/s: 0, 300, 600 [N2O5]/mol L-1: 1.6 x 10-2, 0.8 x 10-2, 0.4 x 10-2
(a) Show that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010, log 4 = 0.6021)
Answer(a) k = (2.303/t) log([A]0/[A])
At 300 s: k = (2.303/300) log(1.6x10-2/0.8x10-2) = 2.31 x 10-3 s-1
At 600 s: k = (2.303/600) log(1.6x10-2/0.4x10-2) = 2.31 x 10-3 s-1
k is constant, therefore it follows first order kinetics.
(b) t1/2 = 0.693/k = 0.693/2.31x10-3 = 300 s
2017 · 3 marks
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021)
AnswerFor 25% decomposition: 20 min = (2.303/k) log(100/75)
For 75% decomposition: t = (2.303/k) log(100/25)
Dividing: 20/t = log(4/3)/log(4) = 0.1250/0.6021
t = (0.6021 x 20)/0.1250 = 96.3 min
2018 · 3 marks
A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 J K-1mol-1)
Answerk2 = 0.693/20, k1 = 0.693/40, so k2/k1 = 2
log(k2/k1) = Ea/(2.303R) x [(1/T1) - (1/T2)]
log 2 = Ea/(2.303 x 8.314) x [(320-300)/(320 x 300)]
Ea = 27663.8 J/mol or 27.66 kJ/mol
2019 · 3 marks
The decomposition of NH3 on platinum surface is zero order reaction. If rate constant k is 4 x 10-3 Ms-1, how long will it take to reduce the initial concentration of NH3 from 0.1M to 0.064M?
AnswerFor zero order reaction: [A] = [A]0 - kt. t = ([A]0 - [A])/k = (0.1 - 0.064)/(4 x 10-3) = 0.036/(4 x 10-3) = 9 s.
2019 · 3 marks
The following data were obtained for reaction A + 2B -> C: Exp 1: [A]=0.2, [B]=0.3, Rate=4.2x10-2. Exp 2: [A]=0.1, [B]=0.1, Rate=6.0x10-3. Exp 3: [A]=0.4, [B]=0.3, Rate=1.68x10-1. Exp 4: [A]=0.1, [B]=0.4, Rate=2.40x10-2.
(a) Find order w.r.t A and B.
(b) Write rate law and overall order.
(c) Calculate rate constant k.
Answer(a) Order w.r.t. A = 2, Order w.r.t. B = 1.
(b) Rate = k[A]2[B]. Overall order = 3.
(c) k = 6.0x10-3/((0.1)2 x 0.1) = 6.0 mol-2 L2 min-1.
2020 · 3 marks
The rate constant for the first order decomposition of N2O5 is given by: k = (2.5 x 1014 s-1) e^(-25000K/T). Calculate Ea for this reaction and rate constant if its half-life period is 300 minutes.
AnswerEa/R = 25000K, so Ea = 25000 x R x K = 25000 x 8.314 = 207850 J/mol = 207.850 kJ/mol. t1/2 = 300 min. k = 0.693/t1/2 = 0.693/300 = 2.31 x 10-3 min-1.
2022-II · 3 marks
Observe the graph shown in figure (log[R0]/[R] vs Time - straight line through origin) and answer the following questions:
(a) What is the order of the reaction?
(b) What is the slope of the curve?
(c) Write the relationship between k and t1/2 (half life period)
Answer(a) The order of the reaction is first order reaction.
(b) Slope of the curve = k / 2.303
(c) Half-life for the first order reaction is given by: t1/2 = 0.693 / k
2023 · 3 marks
(a) For the reaction 2N2O5(g) -> 4NO2(g) + O2(g) at 318 K, calculate the rate of reaction if rate of disappearance of N2O5(g) is 1.4 x 10-3 m s-1.
(b) For a first order reaction derive the relationship t99% = 2t90%.
A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. [R = 8.314 J K-1 mol-1] [Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]
AnswerK1 at 27°C or 300K = 0.693/30 min = 0.0231 min-1 K2 at 47°C or 320K = 0.693/10 min = 0.0693 min-1
Using Arrhenius equation: log(K2/K1) = Ea/(2.303 R) * (T2-T1)/(T1*T2)
log(0.0693/0.0231) = Ea/(2.303 x 8.314 x 10-3 kJ mol-1 K-1) * (20/(300 x 320))
Ea = 43.85 kJ/mol.
2025 · 3 marks
A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction. [R = 8.314 J K-1 mol-1; log 4 = 0.602]
Answert1/2 = 0.693/k. k1 = 0.693/20 = 0.03465 min-1. k2 = 0.693/5 = 0.1386 min-1. log(k2/k1) = Ea/(2.303R) x (1/T1 - 1/T2). log 4 = Ea x 50/(2.303 x 8.314 x 300 x 350). 0.602 = Ea x 50/(2.303 x 8.314 x 300 x 350). Ea = 0.602 x 2.303 x 8.314 x 2100 = 24205.8 J/mol.
2025 · 3 marks
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. [Given: log 4 = 0.602, log 2 = 0.301, R = 8.314 J K-1 mol-1]
Answerk2/k1 = 4. T1 = 293 K, T2 = 313 K. log(k2/k1) = Ea/(2.303R) x (T2-T1)/(T1T2). T1T2 = 293 x 313 = 91709. T2-T1 = 20. log 4 = Ea/(2.303 x 8.314) x 20/91709. 0.602 = Ea x 20/(2.303 x 8.314 x 91709). Ea = 0.602 x 2.303 x 8.314 x 91709/20 = 52854.554 J/mol = 52.85 kJ/mol.
2024 · 3 marks
The rate constant of a reaction quadruples when the temperature changes from 300 K to 320 K. Calculate the activation energy for this reaction. [log 2 = 0.30, log 4 = 0.60, 2.303 R = 19.15 J K-1 mol-1]
AnswerUsing Arrhenius equation: log(k2/k1) = Ea/(2.303R) x (1/T1 - 1/T2)
log 4 = Ea/(2.303 x 8.314) x (1/300 - 1/320)
0.60 = Ea/(19.15) x (20/(300 x 320)) Ea = 0.60 x 2.303 x 8.314 x 300 x 320/20 Ea = 55152 J/mol
2024 · 3 marks
Show that the time required for 99.9% completion in a first order reaction is 10 times of half-life (t1/2) of the reaction. [log 2 = 0.3010, log 10 = 1]
Show that the time required for 99.9% completion in a first order reaction is 10 times of half-life (t1/2) of the reaction. [log 2 = 0.3010, log 10 = 1]
Answerk = 0.693/t1/2. For x = 99.9%: t = (2.303/k) log(a/(a-x)). t99.9%/t50% = (2.303/k log 1000) / (2.303/k log 2) = 3/0.3010 = 10.
Thus t99.9% = 10 x t1/2.
2024 · 3 marks
The rate constant of a reaction quadruples when the temperature changes from 700 K to 720 K. Calculate the activation energy for this reaction. [log 2 = 0.30, log 4 = 0.60, 2.303 R = 19.15 J K-1 mol-1]
AnswerUsing Arrhenius equation: log(k2/k1) = -Ea/(2.303R) x (1/T1 - 1/T2)
log 4 = -Ea/(19.15) x (1/700 - 1/720)
0.60 = -Ea/(19.15) x (20/(700 x 720)) -Ea = 0.60 x 19.15 x 700 x 720/20 Ea = 289548 J/mol = 289.548 kJ/mol
2013 · 3 marks
Calculate the emf of the following cell at 298 K: Fe(s) | Fe2+ (0.001 M) || H+ (1M) | H2(g) (1 bar), Pt(s) (Given E°cell = +0.44 V)
(a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes. (Given: Molar mass of Ag = 108 g mol-1, 1 F = 96500 C mol-1)
(b) Define fuel cell.
Answer(a) m = Zit = (108 x 2 x 15 x 60) / (1 x 96500) = 2.01 g
(b) Fuel cell is the name given to galvanic cells which are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy.
2017 · 3 marks
(a) The cell in which the following reaction occurs: 2Fe3+(aq) + 2I-(aq) → 2Fe2+(aq) + I2(s) has E°cell = 0.236 V at 298 K. Calculate standard Gibb's energy of the cell reaction. (Given: 1 F = 96,500 C mol-1)
(b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1 F = 96,500 C mol-1)
Answer(a) ΔG° = -nFE°cell = -2 x 96500 x 0.236 = -45548 J/mol = -45.55 kJ/mol
(b) Q = It = 0.5 x 2 x 60 x 60 = 3600 C. Number of electrons = (6.023 x 1023 x 3600) / 96500 = 2.25 x 1022 electrons.
2020 · 3 marks
When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO4 and CuSO4 connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A? [Atomic mass: Cu = 63.5 g mol-1, Zn = 65 g mol-1; 1F = 96500 C mol-1]
AnswerFor Cu: n = 2 mol electrons per mol Cu. Q for 2g Cu = (96500 x 2)/63.5 x 2 = 3039.37 C. Time t = Q/I = 3039.37/2 = 1519.68 s = 25 min 33 s. Mass of Zn deposited = 2 x (65/63.5) / (2/2) = 2.0472 g.
2022-II · 3 marks
(a) Calculate delta_r G° and log Kc for the following cell: Ni(s) + 2 Ag+(aq) -> Ni2+(aq) + 2 Ag(s). Given that E° cell = 1.05 V, 1 F = 96,500 C mol-1 OR
(b) Calculate the e.m.f. of the following cell at 298 K: Fe(s) | Fe2+(0.001 M) || H+(0.01 M) | H2(g)(1 bar) | Pt(s). Given that E° cell = 0.44 V. [log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]
(i) Kohlrausch law of independent migration of ions and
(ii) Faraday's first law of electrolysis.
(b) Using E° values of X and Y given below, predict which is better for coating the surface of iron to prevent corrosion and why? Given E°(X2+/X) = -2.36 V, E°(Y2+/Y) = -0.14 V, E°(Fe2+/Fe) = -0.44 V.
Answer(a)
(i) Kohlrausch's law states that the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the limiting molar conductivity of the anions and cations.
(ii) Faraday's First Law of Electrolysis: The mass of a substance deposited at any electrode is directly proportional to the amount of charge passed.
(b) As corrosion is a phenomenon involving the oxidation of iron, it is essential to consider the oxidation potentials. An element with a higher oxidation potential than Fe will oxidise faster than iron preventing corrosion in iron. As E°(X2+/X) = -2.36 V has higher oxidation potential than iron, X can be used for coating the surface of iron.
2025 · 3 marks
Calculate delta_r G° and log Kc of the reaction: Fe2+(aq) + Ag+(aq) -> Fe3+(aq) + Ag(s). Given E°(Ag+/Ag) = 0.80 V, E°(Fe3+/Fe2+) = 0.77 V. [R = 8.314 J K-1 mol-1, F = 96500 C mol-1]
AnswerE°cell = E°cathode - E°anode = 0.80 - 0.77 = 0.03 V. delta_r G° = -nFE°cell = -1 x 96500 x 0.03 = -2895 J/mol = -2.895 kJ/mol. log Kc = nE°cell/0.0591 = 1 x 0.03/0.0591 = 0.507. Hence, delta_r G° and log Kc for the reaction is -2.895 kJ/mol and 0.507 respectively.
2025 · 3 marks
Calculate delta_r G° and log Kc of the reaction: 2Cr(s) + 3Cd2+(aq) -> 2Cr3+(aq) + 3Cd(s). Given E°(Cr3+/Cr) = -0.74 V, E°(Cd2+/Cd) = -0.40 V. [R = 8.314 J K-1 mol-1, F = 96500 C mol-1]
The electrical resistance of a column of 0.05 M NaOH solution of area 0.8 cm2 and length 40 cm is 5 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
AnswerR = 5 x 103 ohm, A = 0.8 cm2 = 0.8 x 10-4 m2, L = 40 cm = 0.4 m. Resistivity rho = R x A/L = (5 x 103 x 0.8 x 10-4)/0.4 = 1.0 ohm m. Conductivity k = 1/rho = 1/1.0 = 1.0 S/m. Molar conductivity = k x 1000/C = 1.0 x 1000/0.05 = 20 S m2/mol.
2025 · 3 marks
Calculate molar conductivity (Lambda_m) for acetic acid and its degree of dissociation (alpha) if its molar conductivity is 48.1 ohm-1 cm2 mol-1. Given that Lambda_m(HCl) = 426 ohm-1 cm2 mol-1, Lambda_m(NaCl) = 126 ohm-1 cm2 mol-1, Lambda_m(CH3COONa) = 91 ohm-1 cm2 mol-1.
(a) Fuel cells are preferred for production of electrical energy than thermal plants.
(b) Iron does not rust even if zinc coating is broken in a galvanised pipe.
(c) In the experimental determination of electrolytic conductance, Direct Current (DC) is not used.
Answer(a) Fuel cells have higher efficiency as they convert chemical energy into electrical energy directly and lower emission while in thermal plants release gases like CO2.
(b) Zinc is more reactive than iron, so it acts as a sacrificial anode that corrodes instead of iron.
(c) When DC is applied, the ions in solution are attracted to the electrodes of opposite charge, leading to their deposition and the concentration of solution will change, so AC current is used which does not alter the concentration of solution.
2024 · 3 marks
Calculate the emf of the cell: Ni(s) + 2Ag+ (0.01 M) -> Ni2+ (0.1 M) + 2Ag(s)
AnswerFrom the cell: Sn -> Sn2+ + 2e- (anode), 2H+ + 2e- -> H2 (cathode) E°cell = E°cathode - E°anode = 0.00 - (-0.14) = -0.14 V (Note: Sn is anode) Applying Nernst Equation: E_cell = E°cell - (0.0592/n) log [Product]/[reactant] E_cell = -0.14 - (0.0592/2) log (0.001/(0.01)2)
= -0.14 - 0.0296 log 10
= -0.14 - 0.0296 x 1
= -0.1696 V
2013 · 3 marks
(a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments? ↑ ↑ ↑ ↑ ↑
(c) What type of point defect is produced when AgCl is doped with CdCl2?
Answer(a) p-type semiconductor is obtained when silicon is doped with boron.
(b) Ferromagnetism is shown when all magnetic moments are aligned in the same direction.
(c) Impurity defect (substitutional solid solution) – when AgCl is doped with CdCl2, Cd2+ replaces two Ag+ ions, creating a cation vacancy.
2015 · 3 marks
An element with molar mass 27 g mol−1 forms a cubic unit cell with edge length 4.05 × 10−8 cm. If its density is 2.7 g cm−3, what is the nature of the cubic unit cell?
Answerd = ZM/(a3NA) Z = d × a3 × NA / M
= 2.7 × (4.05 × 10−8)3 × 6.022 × 1023 / 27
= 2.7 × 66.43 × 10−24 × 6.022 × 1023 / 27
= 4 (approximately)
Since Z = 4, it is a face-centred cubic (FCC) unit cell.
2017 · 3 marks
(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon.
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example.
(b) In AgCl, Ag+ ion is small enough to fit into interstitial sites (large size difference between cation and anion). In NaCl, Na+ and Cl- are of comparable size.
(c) p-type semiconductor (Al has 3 valence electrons, creates electron hole).
2018 · 3 marks
An element 'X' (At. mass = 40 g mol-1) having f.c.c. structure, has unit cell edge length of 400 pm. Calculate the density of 'X' and the number of unit cells in 4 g of 'X'. (NA = 6.022 x 1023 mol-1)
AnswerFor FCC, Z = 4. d = ZM/(a3 x NA) = (4 x 40)/((400 x 10-10)3 x 6.022 x 1023) = 160/(64 x 10-24 x 6.022 x 1023) = 4.15 g cm-3 Number of atoms in 4 g = (4/40) x 6.022 x 1023 = 6.022 x 1022 Number of unit cells = 6.022 x 1022/4 = 1.506 x 1022
2013 · 3 marks
Determine the osmotic pressure of a solution prepared by dissolving 2.5 × 10−2 g of K2SO4 in 2L of water at 25°C, assuming that it is completely dissociated. (R = 0.0821 L atm K−1 mol−1, Molar mass of K2SO4 = 174 g mol−1)
AnswerK2SO4 → 2K+ + SO42− Number of ions produced (i) = 3 π = iCRT = i × (n/V) × R × T π = 3 × (0.025/174) × 0.0821 × 298 / 2
= 3 × 0.025 × 0.0821 × 0.5 × 298 / 174
= 5.27 × 10−3 atm
2015 · 3 marks
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). (Given: Molar mass of benzoic acid = 122 g mol−1, Kf for benzene = 4.9 K kg mol−1)
1.62 = i × 4.9 × (3.9 × 1000)/(122 × 49) 1.62 = i × 4.9 × 3.9 × 1000/(5978) 1.62 = i × 3.196 i = 0.506
Since i < 1, the solute gets associated. Benzoic acid dimerizes in benzene due to intermolecular hydrogen bonding.
2016 · 3 marks
An element crystallizes in a fcc lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 x 1024 atoms.
AnswerFor FCC, Z = 4. Molar mass: M = 300 x 6.022 x 1023 / (2 x 1024) = 90.33 g/mol. a = 250 pm = 2.5 x 10-8 cm. Density = ZM/(a3 x NA) = 4 x 90.33/((2.5 x 10-8)3 x 6.022 x 1023) = 38.4 g/cm3.
2016 · 3 marks
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCl3, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multi-molecular colloids?
Answer(i) Adsorption: Surface phenomenon where molecules accumulate on the surface. Absorption: Bulk phenomenon where molecules are uniformly distributed throughout the body.
(ii) AlCl3 is more effective because Al3+ has higher charge (+3) than Mg2+ (+2). Higher the charge of coagulating ion, greater its coagulating power (Hardy-Schulze rule).
(iii) Sulphur sol forms multi-molecular colloids.
2016 · 3 marks
An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm-3. How many atoms are present in 300 g of the element?
AnswerFor BCC, Z = 2. d = ZM/(a3 x NA). 7.5 = 2M/((500 x 10-10)3 x 6.022 x 1023). M = 7.5 x 75.275/2 = 282.28 g/mol. Number of atoms = (300/282.28) x 6.022 x 1023 = 6.4 x 1023 atoms.
2016 · 3 marks
Define the following terms:
(i) Lyophilic colloid
(ii) Zeta potential
(iii) Associated colloids.
Answer(i) Lyophilic colloid: Colloids in which the dispersed phase has strong affinity for the dispersion medium. They are reversible and self-stabilizing. E.g., starch, gum in water.
(ii) Zeta potential: The potential difference between the fixed charged layer and the diffuse layer of the electrical double layer surrounding a colloidal particle (electrokinetic potential).
(iii) Associated colloids: Substances which behave as normal electrolytes at low concentration but above CMC form micelles and behave as colloids. E.g., soap, detergents.
2016 · 3 marks
Calculate the boiling point of solution when 4 g of MgSO4 (M = 120 g/mol) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization. (Kb for water = 0.52 K kg/mol)
AnswerMgSO4 -> Mg2+ + SO42-, i = 2. delta-Tb = i x Kb x m = 2 x 0.52 x (4 x 1000)/(120 x 100) = 2 x 0.52 x 0.333 = 0.346 K. Boiling point = 373.15 + 0.346 = 373.496 K.
2017 · 3 marks
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1)
(Molar mass of glucose = 180 g mol-1)
AnswerΔTf = (273.15 - 269.15) = 4 K m (sucrose) = (10/342) x (1000/90) = 0.3244 mol kg-1 Kf = ΔTf/m = 4/0.3244 = 12.33 K kg mol-1 m (glucose) = (10/180) x (1000/90) = 0.6166 mol kg-1
ΔTf = 12.33 x 0.6166 = 7.60 K
Tf = 273.15 - 7.60 = 265.55 K
2018 · 3 marks
Give reasons for the following:
(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.
(b) Aquatic animals are more comfortable in cold water than in warm water.
(c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution.
Answer(a) Osmotic pressure is large even for very dilute solutions of macromolecules. Other colligative properties are too small to measure accurately. Also, measurement is at room temperature so proteins don't denature.
(b) Because oxygen is more soluble in cold water (gas solubility decreases with temperature).
(c) Due to dissociation of KCl: KCl(aq) → K+ + Cl-. The van't Hoff factor i is nearly 2, doubling the colligative property.
2019 · 3 marks
(i) What is the role of activated charcoal in gas mask?
(ii) A colloidal sol is prepared by adding FeCl3 solution to NaOH solution. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?
(iii) How does chemisorption vary with temperature?
Answer(i) Adsorption of toxic/poisonous gases.
(ii) Fe2O3.xH2O/OH-. The colloidal particles carry negative charge.
(iii) Chemisorption first increases with increase in temperature then decreases.
2019 · 3 marks
An element crystallizes in fcc lattice with a cell edge of 300 pm. The density of the element is 10.8 g cm-3. Calculate the number of atoms in 108 g of the element.
AnswerFor fcc, Z = 4. Volume of unit cell = (300 x 10-10)3 = 2.7 x 10-23 cm3. Using density = ZM/(NA x a3), M = (10.8 x 2.7 x 10-23 x 6.022 x 1023)/4 = 43.9 g/mol. Number of atoms = (108/43.9) x 6.022 x 1023 = 4.0 x 1024 atoms.
2019 · 3 marks
A 4% solution w/w of sucrose (M = 342 g mol-1) in water has a freezing point of 271.15K. Calculate the freezing point of 5% glucose (M = 180 g mol-1) in water. (Given: Freezing point of pure water = 273.15 K)
AnswerFor sucrose: Delta_Tf = 273.15 - 271.15 = 2.00 K. Molality = (4/342)/(96/1000) = 0.1218 m. Kf = Delta_Tf/m = 2.00/0.1218 = 16.42 K kg mol-1. For glucose: Molality = (5/180)/(95/1000) = 0.2923 m. Delta_Tf = 16.42 x 0.2923 = 4.801 K. Freezing point = 273.15 - 4.801 = 268.35 K.
2019 · 3 marks
(a) Write the dispersed phase and dispersion medium of dust.
(b) Why is physisorption reversible whereas chemisorption is irreversible?
(c) A colloidal sol is prepared by adding KI solution to AgNO3 solution. What is the charge on AgI colloidal particles? How is this sol represented?
Answer(a) Dispersed phase: solid; Dispersion medium: gas (aerosol).
(b) Physisorption involves weak van der Waals forces (reversible); chemisorption involves chemical bond formation (irreversible).
(b) Due to formation of new bonds between adsorbate and adsorbent, surface energy decreases.
(c) At high pressure: x/m = kp0 = k (constant, surface fully saturated).
2019 · 3 marks
An element X with an atomic mass of 81 u has density 10.2 g cm-3. If the volume of unit cell is 2.7 x 10-23 cm3, identify the type of cubic unit cell. (NA = 6.022 x 1023 mol-1)
AnswerZ = (density x V x NA)/M = (10.2 x 2.7x10-23 x 6.022x1023)/81 = 2. Since Z = 2, it is body-centred cubic (BCC).
2019 · 3 marks
A solution containing 1.9 g per 100 mL of KCl (M = 74.5) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60). Calculate the degree of dissociation of KCl solution.
Answerpi(urea) = pi(KCl). (3/60) = i x (1.9/74.5). 0.05 = i x 0.0255. i = 1.96. For KCl (n=2): alpha = (i-1)/(n-1) = (1.96-1)/(2-1) = 0.96 or 96%.
2019 · 3 marks
(a) Write dispersed phase and dispersion medium of butter.
(b) Why does physisorption decrease with increase in temperature?
(c) A colloidal sol is prepared by adding AgNO3 solution to KI solution. What is the charge on AgI colloidal particles? How is this sol represented?
A 0.01 m aqueous solution of AlCl3 freezes at -0.068°C. Calculate the percentage of dissociation. [Given: Kf for Water = 1.68 K kg mol-1]
AnswerGiven m = 0.01 m, Delta Tf = 0.068°C, Kf = 1.86 K kg mol-1. i = Delta Tf / (Kf x m) = 0.068 / (1.86 x 0.01) = 3.65. AlCl3 -> Al3+ + 3Cl-. i = 1 + 3*alpha. 3.65 = 1 + 3*alpha. alpha = (3.65-1)/3 = 0.883. Percentage dissociation = 88.3%.
2020 · 3 marks
Write three differences between lyophobic sol and lyophilic sol.
OR Define the following terms:
(i) Protective colloid
(ii) Zeta potential
(iii) Emulsifying agent
AnswerLyophilic sols: self-stabilizing, reversible, high viscosity. Lyophobic sols: need stabilizing agents, irreversible, low viscosity. OR
(i) Protective colloid: A lyophilic colloid that protects lyophobic colloid from coagulation.
(ii) Zeta potential: The potential difference between the fixed layer and the diffused layer of opposite charges.
(iii) Emulsifying agent: A substance that stabilizes an emulsion, e.g. soap.
2020 · 3 marks
The freezing point of a solution containing 5g of benzoic acid (M = 122 g mol-1) in 35g of benzene is depressed by 2.94 K. What is the percentage association of benzoic acid if it forms a dimer in solution? (Kf for benzene = 4.9 K kg mol-1)
AnswerMB = (Kf x WB)/(WA x Delta Tf) = (4.9 x 5)/(0.035 x 2.94) = 24.5/0.1029 = 238 g/mol. i = normal molar mass / observed molar mass = 122/238 = 0.513. For dimerization: n = 2, alpha = (i-1)/(1/n - 1) = (0.513-1)/(0.5-1) = (-0.487)/(-0.5) = 0.974. Percentage association = 97.4%.
2023 · 3 marks
When 19.5 g of F-CH2-COOH (Molar mass = 78 g mol-1), is dissolved in 500 g of water, the depression in freezing point is observed to be 1°C. Calculate the degree of dissociation of F-CH2-COOH. [Given: Kf for water = 1.86 K kg mol-1]
AnswerMolecular mass of 78 g/mol (F-CH2COOH). No. of moles of fluoroacetic acid = 19.5/78 = 0.25. Molality = 0.25/(500/1000) = 0.50 m. Delta_Tf = Kf x m = 1.86 x 0.50 = 0.93 K. i = 1.0/0.93 = 1.0753. CH2FCOOH -> CH2FCOO- + H+. C(1-alpha) + C*alpha + C*alpha = C(1+alpha). i = C(1+alpha)/C = 1+alpha = 1.0753. alpha = 0.0753.
(ii) alpha = 0.50 x 0.0753 = 0.03765. C(1-alpha) = 0.50(1-0.0753) = 0.462.
2025 · 3 marks
A solution of glucose (molar mass = 180 g mol-1) in water has a boiling point of 100.20 C. Calculate the freezing point of the same solution. Molal constants for water Kf and Kb are 1.86 K kg mol-1 and 0.512 K kg mol-1 respectively.
AnswerTb of glucose solution = 100.20 C, delta Tb = 100.20 - 100 = 0.20 C. delta Tb = Kb x m, m = 0.20/0.512 = 0.390 mol/kg. delta Tf = Kf x m = 1.86 x 0.390 = 0.725 C. Freezing point of solution = 0 - 0.725 = -0.725 C.
2025 · 3 marks
A solution containing 15 g urea (molar mass = 60 g mol-1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 180 g mol-1) in water. Calculate the mass of glucose present in one litre of its solution.
AnswerFor isotonic solutions, pi_urea = pi_glucose. Since pi = CRT and RT is common, C_urea = C_glucose. Moles of urea = 15/60 = 0.25 mol for 1 litre. n_glucose = 0.25 mol. Mass of glucose = 0.25 x 180 = 45 g. Hence, the mass of glucose present in one litre of its solution is 45 g.
2025 · 3 marks
Henry's law constant for CO2 in water is 1.67 x 108 Pa at 298 K. Calculate the number of moles of CO2 in 500 ml of soda water when packed under 2.53 x 105 Pa at the same temperature.
AnswerUsing Henry's law: P_CO2 = K_H x x_CO2. x_CO2 = 2.53 x 105 / 1.67 x 108 = 1.51 x 10-3. Mass of water in 500 mL = 500 g. n_H2O = 500/18 = 27.78 mol. x_CO2 = n_CO2/(n_CO2 + n_H2O). 1.51 x 10-3 = n_CO2/(n_CO2 + 27.78). n_CO2 = 1.51 x 10-3 x 27.78 = 0.042 mol.
2025 · 3 marks
At 25 C the saturated vapour pressure of water is 24 mm Hg. Find the saturated vapour pressure of a 5% aqueous solution of urea at the same temperature. (Molar mass of urea = 60 g mol-1)
AnswerMass of urea = 5% of 1000 g = 50 g. Moles of urea = 50/60 = 0.8333 mol. Mass of water = 1000 - 50 = 950 g. Moles of water = 950/18 = 52.7778 mol. Mole fraction of water = 52.7778/(52.7778 + 0.8333) = 0.9844. P solution = P_water x X_water = 24 mm Hg x 0.9844 = 23.634 mm Hg.
2025 · 3 marks
Henry's law constant for CO2 in water is 1.67 x 108 Pa at 298 K. Calculate the number of moles of CO2 in 540 g of soda water when packed under 3.34 x 105 Pa at the same temperature.
Answerp = K_H x chi. chi = p/K_H = 3.34 x 105 / 1.67 x 108 = 2 x 10-3. N = 540/18 = 30. n_CO2 = chi x N_H2O = 2 x 10-3 x 30 = 60 x 10-3 = 0.06 moles.
2014 · 3 marks
(a) In reference to Freundlich adsorption isotherm, write the expression for adsorption of gases on solids in the form of an equation.
(b) Write an important characteristic of lyophilic sols.
(c) Based on type of particles of dispersed phase, give one example each of associated colloid and multimolecular colloid.
Answer(a) x/m = kP1/n (where x = mass of adsorbate, m = mass of adsorbent, P = pressure, k and n are constants, 1/n is between 0 and 1)
(b) Lyophilic sols are reversible in nature. They are self-stabilizing and do not need stabilizing agents.
(c) Associated colloid: Soap (sodium stearate) forms micelles. Multimolecular colloid: Gold sol (Au atoms aggregate).
2015 · 3 marks
Give reasons for the following observations:
(i) Leather gets hardened after tanning.
(ii) Lyophilic sol is more stable than lyophobic sol.
(iii) It is necessary to remove CO when ammonia is prepared by Haber's process.
Answer(i) Tanning involves cross-linking of collagen fibres with chromium salts, which makes the leather harder and more durable.
(ii) Lyophilic sols are self-stabilizing because the colloidal particles are solvated (have a layer of solvent molecules). They do not need extra stabilizing agents, unlike lyophobic sols.
(iii) CO acts as a poison for the iron catalyst used in Haber's process. It must be removed to maintain catalyst efficiency.
2017 · 3 marks
Write one difference in each of the following:
(i) Lyophobic sol and lyophilic sol
(ii) Solution and Colloid
(iii) Homogeneous catalysis and heterogeneous catalysis.
Answer(i) Lyophobic sols are irreversible and less stable; lyophilic sols are reversible and more stable.
(ii) Solutions have particle size < 1 nm; colloids have particle size 1-1000 nm.
(iii) In homogeneous catalysis, catalyst is in same phase as reactants; in heterogeneous catalysis, catalyst is in different phase.
2017 · 3 marks
Write one difference between each of the following:
(i) Multimolecular colloid and Macromolecular colloid
(ii) Sol and Gel
(iii) O/W emulsion and W/O emulsion
Answer(i) Multimolecular colloids: aggregate of atoms/molecules (size < 1 nm) e.g., gold sol. Macromolecular colloids: single large molecules (size in colloidal range) e.g., starch.
(ii) Sol: solid dispersed in liquid. Gel: liquid dispersed in solid.
(iii) O/W emulsion: oil dispersed in water (e.g., milk). W/O emulsion: water dispersed in oil (e.g., butter).
2017 · 3 marks
Write one difference in each of the following:
(a) Multimolecular colloid and associated colloid
(b) Coagulation and peptization
(c) Homogeneous catalysis and heterogeneous catalysis
OR
(a) Write the dispersed phase and dispersion medium of milk.
(b) Write one similarity between physisorption and chemisorption.
(c) Write the chemical method by which Fe(OH)3 sol is prepared from FeCl3.
Answer(a) Multimolecular: aggregate of many small molecules. Associated: aggregates of surfactant molecules (micelles) above CMC.
(b) Coagulation: settling of colloidal particles. Peptization: conversion of precipitate into colloidal sol.
(c) Homogeneous: catalyst in same phase. Heterogeneous: catalyst in different phase.
(a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution?
(b) persistent dialysis of a colloidal solution is carried out?
(c) an emulsion is centrifuged?
Answer(a) A colloidal sol of Fe(OH)3 is formed (peptization). FeCl3 provides Fe3+ ions which are adsorbed on Fe(OH)3 particles.
(b) The colloidal solution may become unstable and coagulate as the stabilizing electrolyte is completely removed.
(c) The emulsion breaks into two separate layers of constituent liquids (demulsification).
2025 · 4 marks
The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reactions is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction. Answer the following: (a)
(i) What is a rate determining step?
(ii) Define complex reaction.
(b) What is the effect of temperature on the rate constant of a reaction?
OR
(b) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions?
(c) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y?
Answer(a)
(i) The rate determining step is the slowest step of a chemical reaction that determines the speed at which the overall reaction proceeds.
(ii) When overall reaction does not occur in a single step, but occurs in multiple steps, such that order is not equal to the molecularity of the reaction. The mechanism of a complex reaction consists of more than one elementary step.
(b) With increase in temperature, the rate of the reaction and the rate constant increases.
OR
(b) A complex reaction proceeds through several elementary reactions. The number of molecules involved in each elementary reaction may be different, i.e., the molecularity of each step may be different. Therefore, it is impossible to determine the molecularity of overall complex reaction. Order of a complex reaction is determined by the slowest step in its mechanism.
(c) For second order, Rate = k[X]2. When [X] is increased 3 times, Rate = k(3a)2 = 9ka2. Thus, the rate of formation of Y will become 9 times.
2023 · 4 marks
Rahul set-up an experiment to find resistance of aqueous KCl solution for different concentrations at 298 K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with a.c. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from null point he also calculated the conductivity K and molar conductivity Lambda_m and recorded his readings in tabular form. S.No. | Conc. (M) | k S cm-1 | Lambda_m S cm2 mol-1
1. | 1.00 | 111.3 x 10-3 | 111.3
2. | 0.10 | 12.9 x 10-3 | 129.0
3. | 0.01 | 1.41 x 10-3 | 141.0
Answer the following questions:
(a) Why does conductivity decrease with dilution?
(b) If Lambda_m° of KCl is 150.0 S cm2 mol-1, calculate the degree of dissociation of 0.01 M KCl.
(c) If Rahul had used HCl instead of KCl then would you expect the Lambda_m values to be more or less than those per KCl for a given concentration. Justify.
Answer(a) Conductivity decreases with dilution because it depends upon the number of ions present in the solution. When dilution increases number of available ions decreases. Hence, conductivity decreases.
(b) alpha = Lambda_m / Lambda_m°
Lambda_m = 141.0 S cm2 mol-1 Lambda_m° = 150.0 S cm2 mol-1 alpha = 141/150 = 0.94
(c) Molar conductivity of HCl will be high because when it breaks down into ions, it produces HCl -> H+ + Cl-. Its H+ (cation) size is smaller than K+ ion. So for same concentration of HCl and KCl, HCl shows high molar conductivity.
2023 · 4 marks
32 (OR) (c): Amit, a classmate of Rahul repeated the same experiment with CH3COOH solution instead of KCl solution. Give one point that would be similar and one that would be different in his observations as compared to Rahul.
Answer(i) KCl is strong electrolyte and completely dissociate into their respective ion while CH3-COOH is weak electrolyte and do not completely dissociate.
(ii) Number of ions produced after dissociation are equal: KCl -> K+ + Cl- and CH3-COOH -> CH3-COO- + H+. Similar point: Both produce same number of ions on dissociation. Different point: KCl is a strong electrolyte (fully dissociates) while CH3COOH is a weak electrolyte (partially dissociates), so Lambda_m values for CH3COOH will be lower.
2024 · 4 marks
In a galvanic cell, chemical energy of a redox reaction is converted into electrical energy, whereas in an electrolytic cell the redox reaction occurs on passing electricity. Based on this:
(a) What is the function of a salt bridge in a galvanic cell?
(b) When does galvanic cell behave like an electrolytic cell?
(c) Can copper sulphate solution be stored in a pot made of zinc? Explain with the help of the value of E° cell. (E° Cu2+/Cu = 0.34 V, E° Zn2+/Zn = -0.76 V)
OR
(c) How much charge in terms of Faraday is required for the following:
(i) 1 mol of MnO4- to Mn2+
(ii) 1 mol of H2O to O2
Answer(a) It maintains the electrical neutrality and completes the circuit.
(b) When the external opposite potential is applied greater than the voltage of the reaction, the reaction starts proceeding in reverse direction. At this point the cell becomes an electrolytic cell.
(c) No, copper sulphate in zinc pot will have a spontaneous substitution reaction. CuSO4 + Zn -> Cu + ZnSO4. E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 1.10 V. Since E°cell is positive, the reaction is spontaneous.
OR (c)
(i) MnO4-(+7) -> Mn2+. Change in oxidation number = 7-2 = 5. Faraday's required = 5F.
(ii) H2O -> O2. Change in oxidation number = 2. Faraday's required = 2F.
2025 · 4 marks
The spontaneous flow of the solvent through a semipermeable membrane from a pure solvent to a solution or from a dilute solution to a concentrated solution is called osmosis. [Passage about osmosis experiment with eggs]. Answer the following:
(a) Define reverse osmosis. Name one SPM which can be used in the process of reverse osmosis. (b)
(i) What do you expect to happen when red blood corpuscles (RBCs) are placed in 0.5% NaCl solution?
OR (b)
(ii) Which one of the following will have higher osmotic pressure in 1 M KCl or 1 M urea solution. Justify your answer.
(c) Why osmotic pressure is a colligative property?
Answer(a) Reverse Osmosis (RO) is a water purification process that uses a semi-permeable membrane to remove impurities. In reverse osmosis, water is forced through the membrane under pressure against the natural osmotic flow. One of the most common materials used for semi-permeable membranes is Thin Film Composite (TFC) membranes. (b)
(i) RBCs placed in 0.5% NaCl solution will swell and may undergo haemolysis (bursting), as water moves into the cells due to osmosis.
OR (b)
(ii) 1 M KCl will have higher osmotic pressure. For KCl, i = 2 (it dissociates into K+ and Cl-). For urea, i = 1. Since pi = iCRT, osmotic pressure of KCl will be higher.
(c) Osmotic pressure is a colligative property because it depends only on the number of solute particles in the solution, not on the nature or identity of the solute particles themselves.
2013 · 5 marks
(a) A reaction is second order in A and first order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of A three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled?
(b) A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2 of this reaction. (Given log 1.428 = 0.1548)
Answer(a)
(i) Rate = k[A]2[B]
(ii) If [A] increases 3 times: Rate2/Rate1 = (3a)2 × b / (a2 × b) = 9. Rate becomes 9 times.
(iii) If [A] and [B] are doubled: Rate2/Rate1 = (2a)2 × (2b) / (a2 × b) = 8. Rate becomes 8 times.
(a) For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
(b) Rate constant 'k' of a reaction varies with temperature 'T' according to the equation:
log k = log A − Ea/(2.303R) × (1/T)
Where Ea is the activation energy. When a graph is plotted for log k Vs. 1/T, a straight line with a slope of −4250 is obtained. Calculate Ea for the reaction. (R = 8.314 JK−1 mol−1)
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained: t/s: 0, 30, 60 [CH3COOCH3]/mol L−1: 0.60, 0.30, 0.15
(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(ii) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(Given log 2 = 0.3010, log 4 = 0.6021)
Answer(i) k = (2.303/t) log([A0]/[A])
At t = 30: k = (2.303/30) log(0.60/0.30) = (2.303/30) × 0.3010 = 0.023 s−1
At t = 60: k = (2.303/60) log(0.60/0.15) = (2.303/60) × 0.6021 = 0.023 s−1
Since k is constant in both readings, it is a pseudo-first order reaction.
(ii) Average rate = −Δ[R]/Δt = −(0.15 − 0.30)/(60 − 30) = 0.15/30 = 0.005 mol L−1 s−1
2015 · 5 marks
(a) For a reaction A + B → P, the rate is given by Rate = k[A][B]2
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010)
Answer(a)
(i) If [B] is doubled: Rate2 = k[A](2[B])2 = 4k[A][B]2 = 4 × Rate1. Rate increases 4 times.
(ii) If A is in large excess, [A] is essentially constant. The reaction becomes pseudo second order (order with respect to B = 2).
(b) t1/2 = 30 min, k = 0.693/30 = 0.0231 min−1
For 90% completion: t = (2.303/k) log(100/10) = (2.303/0.0231) × 1 = 99.7 min
2020 · 5 marks
(a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?
(b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics. [Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021, log 5 = 0.6991] OR
(a) A first order reaction is 50% completed in 30 minutes at 300K and in 10 minutes at 320K. Calculate activation energy (Ea) for the reaction. (R = 8.314 J K-1 mol-1)
(b) Write the two conditions for collisions to be effective collisions.
(c) How order of reaction and molecularity differ towards a complex reaction?
Answer(a) t = (2.303/k) log([A]0/[A]). 40 = (2.303/k) log(100/75). k = (2.303/40) x log(4/3) = (2.303/40)(0.6021-0.4771) = (2.303 x 0.125)/40 = 7.196 x 10-3 min-1. For 80% completion: t = (2.303/k) log(100/20) = (2.303/7.196x10-3) x log 5 = (2.303 x 0.6991)/0.007196 = 223.7 min.
(b) Order of reaction is the sum of the coefficients of the reacting species in the rate equation. A bimolecular reaction follows first order kinetics when one of the reactants is taken in large excess.
OR
(a) k1 = 0.693/30 = 0.0231 min-1, k2 = 0.693/10 = 0.0693 min-1. Using Arrhenius equation: log(k2/k1) = Ea/(2.303R) x (T2-T1)/(T1T2). Ea = 43.85 kJ/mol.
(b) Reactant molecules must have sufficient energy to break bonds; molecules must have proper orientation.
(c) For complex reactions, molecularity has no significance while order of reaction is applicable.
2022-II · 5 marks
Case-based passage on Chemical Kinetics: The rate of reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst affect the rate of reaction. Mathematical representation of rate of reaction is given by rate law: Rate = k[A]^x[B]^y. x and y indicate how sensitive the rate is to change in concentration of A and B. Sum of x + y gives the overall order of a reaction. When a sequence of elementary reaction gives us the products, the reaction is called complex reaction. Molecularity and order of an elementary reaction are same. Zero order reactions are relatively uncommon but they occur under special condition. All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
(a) What is the effect of temperature on the rate constant of a reaction? (b) For a reaction A + B -> Product, the rate given by rate = k[A]2[B]1/2. What is the order of the reaction? (c) How order and molecularity are different for complex reactions? (d) A first order reaction has a rate constant 2 x 10-3 s-1. How long will 6 g of this reactant take to reduce to 2g.
OR
(d) The half life for radioactive decay of 14C is 6930 years. An archaeological artifact containing wood had only 75% of the 14C found in a living tree. Find the age of the sample. [log 4 = 0.6021, log 3 = 0.4771, log 2 = 0.3010, log 10 = 1]
Answer(a) The rate constant (k) for a reaction increases with increase in temperature and becomes almost double with every 10° rise in temperature. This effect is expressed by Arrhenius equation: k = Ae^(-Ea/RT).
(b) According to the equation, r = k[A]2[B]1/2. Order of the reaction = 2 + 1/2 = 5/2.
(c) Order of reaction is defined as the sum of the powers of the molar concentration of the reaction species in the rate equation. It is applicable for both elementary and complex reactions. Molecularity of reaction is defined as the total number of reacting species participating in an elementary reaction. It has no significance for complex reactions as applicable for only elementary reactions.
(d) k = 2 x 10-3 s-1. t = (2.303/k) * log([R]0/[R]) = (2.303/(2 x 10-3)) * log(6/2) = 1151.5 * 0.4771 = 549.3 s.
OR
(d) t1/2 = 6930 years (note: some sources use 6980). k = 0.693/6930. Initial concentration [R]0 = 100, [R] = 75. t = (2.303/k) * log(100/75) = (2.303/(0.693/6930)) * log(4/3) = (2.303 * 6930/0.693) * 0.1249 = approximately 2898 years.
2014 · 5 marks
(a) Define the following terms:
(i) Limiting molar conductivity
(ii) Fuel cell
(b) Resistance of a conductivity cell filled with 0.1 mol L−1 KCl solution is 100 Ω. If the resistance of the same cell when filled with 0.02 mol L−1 KCl solution is 520 Ω, calculate the conductivity and molar conductivity of 0.02 mol L−1 KCl solution. The conductivity of 0.1 mol L−1 KCl solution is 1.29 × 10−2 Ω−1 cm−1.
Answer(a)
(i) When the concentration of electrolyte approaches zero, the molar conductivity is termed as limiting molar conductivity (Λ0m).
(ii) Fuel cells are galvanic cells that transform chemical energy into electrical energy from fuel combustion by redox reaction, such as hydrogen, methanol, etc.
(b) Cell constant (G*) = k × R = 1.29 × 10−2 × 100 = 1.29 cm−1
For 0.02 M KCl: k = G*/R = 1.29/520 = 2.48 × 10−3 Ω−1 cm−1
(a) State Faraday's first law of electrolysis. How much charge in terms of Faraday is required for the reduction of 1 mol of Cu2+ to Cu.
(b) Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) | Cu2+(0.01 M) | Cu(s) [Give E0cell = ± 2.71 V, F = 96500 C mol−1]
Answer(a) Faraday's first law: The mass of substance deposited at any electrode is directly proportional to the amount of charge passed (m = Z × Q). Cu2+ + 2e− → Cu. Since 2 mol of electrons are involved, 2F charge is required.
(b) Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)
Ecell = E°cell − (2.303RT/nF) log([Mg2+]/[Cu2+])
= 2.71 − (0.0591/2) × log(0.1/0.01)
= 2.71 − (0.0591/2) × log 10
= 2.71 − 0.02955 = 2.68 V
2016 · 5 marks
(a) Calculate E°cell for the following reaction at 298K: 2Al(s) + 3Cu2+(0.01M) -> 2Al3+(0.01M) + 3Cu(s). Given: E°cell = 1.98 V.
(b) Using E° values of A and B, predict which is better for coating iron [E°(Fe2+/Fe) = -0.44V] to prevent corrosion. Given: E°(A2+/A) = -2.37 V, E°(B2+/B) = -0.14 V.
OR
(a) The conductivity of 0.001 mol/L solution of CH3COOH is 3.905 x 10-5 S/cm. Calculate molar conductivity and degree of dissociation. Given lambda0(H+) = 349.6, lambda0(CH3COO-) = 40.9 S cm2/mol.
(b) Define electrochemical cell. What happens if external potential exceeds E°cell?
(b) Metal A (E° = -2.37 V) is better as it is more negative than Fe; it gets oxidized preferentially (sacrificial protection).
OR
(a) Lambda_m = 3.905 x 10-5 x 1000/0.001 = 39.05 S cm2/mol. Lambda0 = 349.6 + 40.9 = 390.5. alpha = 39.05/390.5 = 0.1 (10%).
(b) Electrochemical cell produces electricity from spontaneous reactions. When external potential exceeds E°cell, reaction reverses (becomes electrolytic cell).
2018 · 5 marks
(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: Sn(s) | Sn2+(0.004 M) || H+(0.020 M) | H2(g)(1 bar) | Pt(s) (Given: E°Sn2+/Sn = -0.14 V)
(b) Give reasons:
(i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH3COOH decreases on dilution.
OR
(a) For the reaction: 2AgCl(s) + H2(g)(1 atm) → 2Ag(s) + 2H+(0.1 M) + 2Cl-(0.1 M), ΔG° = -43600 J at 25°C. Calculate the e.m.f. of the cell. [log 10-n = -n]
(b) Define fuel cell and write its two advantages.
(i) Due to overpotential/overvoltage of O2, Cl2 is liberated at lower applied voltage.
(ii) Conductivity decreases because number of ions per unit volume decreases on dilution.
OR
(a) ΔG° = -nFE°. -43600 = -2 x 96500 x E°. E° = 0.226 V
E = 0.226 - (0.059/2) log[(0.1)2 x (0.1)2/1] = 0.226 - 0.059/2 x log 10-4 = 0.226 + 0.118 = 0.344 V
(b) Fuel cells convert energy of combustion of fuels directly into electrical energy. Advantages: High efficiency, non-polluting.
2019 · 5 marks
E°cell for the given redox reaction is 2.71 V. Mg + Cu2+(0.01M) -> Mg2+(0.001M) + Cu. Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) When external voltage < 2.71 V: electrons flow from anode to cathode (Mg to Cu), same direction.
(ii) When external voltage > 2.71 V: reaction is reversed, current flows in opposite direction (Cu to Mg).
2019 · 5 marks
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass of Zn deposited. (Molar mass: Fe = 56, Zn = 65.3, 1F = 96500 C mol-1)
(b) In the plot of molar conductivity vs. square root of concentration, following curve obtained for two electrolytes A and B:
(i) Predict the nature of electrolytes A and B.
(ii) What happens on extrapolation of Lambda_m to concentration approaching zero for electrolytes A and B?
Answer(a) m = zIt/F. 2.8 = (56 x 2 x t)/(2 x 96500). t = 4825 s. Mass of Zn = 2.8 x (65.3/2)/(56/2) = 3.265 g. (b)
(i) A is strong electrolyte (linear), B is weak electrolyte (curved).
(ii) For strong electrolyte A, Lambda_m0 can be obtained by extrapolation. For weak electrolyte B, Lambda_m0 cannot be obtained by extrapolation as curve rises steeply.
2019 · 5 marks
(a) Conductivity of 0.001 mol/L acetic acid is 4.95x10-5 S cm-1. Calculate dissociation constant if Lambda_m0 for acetic acid is 390.5 S cm2 mol-1.
(b) Strong: Lambda_m decreases slightly with concentration. Weak: Lambda_m increases sharply at low concentration. Lambda_m0 for weak electrolyte obtained by Kohlrausch's law: Lambda_m0 = v+*lambda_+0 + v-*lambda_-0.
2020 · 5 marks
(a) The electrical resistance of a column of 0.05 M KOH solution of length 50 cm and area of cross-section 0.625 cm2 is 5 x 103 ohm. Calculate its resistivity, conductivity and molar conductivity.
(b) Predict the products of electrolysis of an aqueous solution of CuCl2 with platinum electrodes. [Given: E°(Cu2+/Cu) = +0.34V, E°(1/2 Cl2/Cl-) = +1.36V, E°(H+/H2) = 0.00V, E°(1/2 O2/H2O) = +1.23V] OR
(a) Calculate the e.m.f. of the cell: Zn(s)/Zn2+(0.1M) || (0.01M)Ag+/Ag(s). [Given: E°(Zn2+/Zn) = -0.76V, E°(Ag+/Ag) = +0.80V, log 10 = 1]
(b) X and Y are two electrolytes. On dilution molar conductivity of X increases 2.5 times while that of Y increases 25 times. Which of the two is a weak electrolyte and why?
Answer(a) Cell constant = l/A = 50/0.625 = 80 cm-1. Resistivity = R x A/l = 5000 x 0.625/50 = 62.5 ohm cm. Conductivity = 1/resistivity = 1/62.5 = 0.016 S cm-1. Molar conductivity = (K x 1000)/M = (0.016 x 1000)/0.05 = 320 S cm2 mol-1.
(b) At cathode: Cu2+ + 2e- -> Cu(s) (E° = 0.34V, higher than H+). At anode: Cl- gets oxidised to Cl2 gas (due to overpotential of O2).
(b) Y is a weak electrolyte. On dilution, weak electrolyte's molar conductivity increases sharply (25 times) due to increased dissociation, while strong electrolyte (X) shows only modest increase (2.5 times).
2022-II · 5 marks
Case-based passage on Electrochemistry: Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. A salt bridge also connects to the half cells. If E°cell is positive the reaction is spontaneous and if negative it is non-spontaneous (electrolytic cell). Electrolysis refers to decomposition of a substance by an electric current. Conductivity is represented by k and depends upon nature and concentration of electrolyte. Molar conductivity is conductance of the volume of solution containing one mole of electrolyte. Limiting molar conductivity of weak electrolytes cannot be obtained graphically.
(a) Is silver plate the anode or cathode? (b) What will happen if the salt bridge is removed? (c) When does electrochemical cell behave like an electrolytic cell? (d)
(i) What will happen to the concentration of Zn2+ and Ag+ when Ecell = 0? (ii) Why does conductivity of a solution decrease with dilution?
OR
(d) The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2 mol-1. Calculate the conductivity of this solution.
Answer(a) Silver plate acts as cathode.
(b) Salt bridge permits the flow of current by completing the circuit as well as it maintains the charge balance between anode and cathode by movement of electrons. If the salt bridge is removed no current will flow in the circuit and the voltage will drop to zero.
(c) An electrochemical cell behaves like an electrolytic cell when there is an application of an external opposite potential on the galvanic cell and reaction is not inhibited until the opposing voltage reaches the value 1.1 V. At this stage, no current flows through the cell and on increasing the external potential any further the reaction will function in the opposite direction. Eext > Ecell.
(d)
(i) When Ecell = 0, an equilibrium condition is reached and the concentration of Zn2+ and Ag+ remains same.
(ii) Conductivity of a solution defined as the conductance of ions present in a unit volume of the solution. On dilution, the number of ions per unit volume decreases. Thus, the conductivity of the solution decreases on dilution.
OR
(d) Molar Conductivity = Conductivity / Concentration. Conductivity = molar conductivity * C = (138.9 S cm2 mol-1 * 1.5 mol/L) / (1000 cm3/L) = 0.208 S cm-1.
2023 · 5 marks
(a) Conductivity of 2 x 10-3 M methanoic acid is 8 x 10-5 S cm-1. Calculate its molar conductivity and degree of dissociation if Lambda_m0 for methanoic acid is 404 S cm2 mol-1.
Answer(a) Molar conductivity: Lambda_m = (kappa x 1000)/C = (8 x 10-5 S cm-1 x 1000)/(2 x 10-3 mol L-1) = (8 x 10-2)/(2 x 10-3) = 40 S cm2 mol-1. Degree of dissociation: alpha = Lambda_m / Lambda_m0 = 40/404 = 0.099
2023 · 5 marks
(b) Calculate the Delta_rG0 and log Kc for the given reaction at 298 K: Ni(s) + 2Ag+(aq) <=> Ni2+(aq) + 2Ag(s)
Given: E0_Ni2+/Ni = -0.25 V, E0_Ag+/Ag = +0.80 V
1F = 96500 C mol-1.
Answer(b) Ni -> Ni2+ (E0 = -0.25V) (Oxidation half) 2Ag+ -> 2Ag (E0 = 0.80V) (Reduction half) E° = Ec - Ea = 0.80 - (-0.25) = 1.05 V Delta_G° = -nFE° = -2 x 96500 x 1.05 = -202,650 J mol-1 = -202.65 kJ mol-1
log Kc = (nE°)/0.0591 = (1.05 x 2)/0.0591 = 35.53
Kc = Antilog(35.53) = 3.38 x 1035
2025 · 5 marks
(A)(a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: Sn(s)|Sn2+(0.004M)||H+(0.02M)|H2(g)(1 Bar)|Pt(s). (Given: E°(Sn2+/Sn) = -0.14 V, E°(H+|H2(g),Pt) = 0.00V).
(b) Account for the following:
(i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH3COOH decreases on dilution.
OR (B)(a) Write the anode and cathode reaction and the overall cell reaction occurring in a lead storage battery during its use.
(b) Calculate the potential for half-cell containing 0.01 M K2Cr2O7(aq), 0.01 M Cr3+(aq) and 1.0 x 10-4 M H+(aq). The half cell reaction is Cr2O72-(aq) + 14H+(aq) + 6e- -> 2Cr3+(aq) + 7H2O(l) and the standard electrode potential is given as E° = 1.33 V. [Given: log 10 = 1]
Answer(A)(a) At anode: Sn -> Sn2+ + 2e-. At cathode: 2H+ + 2e- -> H2. E°cell = E°SHE - E°(Sn2+/Sn) = 0 - (-0.14) = +0.14 V. Ecell = E°cell - 0.0591/2 log [Sn2+] x PH2/[H+]2 = 0.14 - 0.0591/2 log 0.004 x 0.987/[0.020]2 = 0.111V. (b)
(i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl. It is due to the phenomenon of over-voltage. The over-voltage required for formation of oxygen is much larger than that required for the formation of chlorine.
(ii) Acetic acid (CH3COOH) is a weak electrolyte. During dilution, the number of ions per unit volume decreases, despite an increase in the degree of ionization. Since the number of free ions affects conductivity, dilution leads to a decrease in overall conductivity.
(A)(a) Calculate the standard Gibbs energy (delta_r G°) of the following reaction at 25 C: Au(s) + Ca2+(1M) -> Au3+(1M) + Ca(s). E°(Au3+/Au) = +1.5 V, E°(Ca2+/Ca) = -2.87 V. Predict whether the reaction will be spontaneous or not at 25 C. [1 F = 96500 C mol-1].
(b) Tarnished silver contains Ag2S. Can this tarnish be removed by placing tarnished silverware in an aluminium pan containing an inert electrolytic solution such as NaCl? The standard electrode potential for half reaction: Ag2S(s) + 2e- -> 2Ag(s) + S2- is -0.71 V and for Al3+ + 3e- -> 2Al(s) is -1.66 V.
OR (B)(a) Define the following:
(i) Cell potential
(ii) Fuel cell.
(b) Calculate emf of the following cell at 25 C: Zn(s)(0.1M) | Zn2+(0.1M) || Cd2+(0.01M) | Cd(s). Given: E°(Cd2+/Cd) = -0.40 V, E°(Zn2+/Zn) = -0.76 V. [log 10 = 1]
Answer(A)(a) E°cell = E°cathode - E°anode = (-2.87) - (1.5) = -4.37 V. n = 6 (LCM of 3 and 2). delta_r G° = -nFE°cell = -6 x 96500 x (-4.37) = +2530230 J/mol = +2530.23 kJ/mol. Since delta_r G° is positive, the reaction is non-spontaneous at 25 C.
(b) Yes, tarnished silver can be removed. Aluminium (E° = -1.66 V) has a more negative electrode potential than Ag2S/Ag (E° = -0.71 V), so Al acts as the reducing agent.
OR (B)(a)
(i) Cell potential (electromotive force or emf) is the difference in potential between the two electrodes (cathode and anode) in a galvanic or electrolytic cell.
(ii) A fuel cell is an electrochemical device that converts the chemical energy of a fuel (typically hydrogen) and an oxidising agent (typically oxygen) directly into electrical energy, water and heat.
(a) An element has atomic mass 93 g mol-1 and density 11.5 g cm-3. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(b) Write any two differences between amorphous solids and crystalline solids.
OR
(a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol-1)
(b) Give reasons:
(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.
(ii) Silicon on doping with phosphorus forms n-type semiconductor.
(iii) Ferrimagnetic substances show better magnetism than antiferromagnetic substances.
Answer(a) Using d = ZM/(a3NA), Z = (d x a3 x NA)/M = (11.5 x (300x10-10)3 x 6.022x1023)/93 = 2. BCC unit cell.
(b) Crystalline solids have definite geometry, sharp melting point; amorphous solids have irregular shape, melt over a range.
OR
(a) Moles of Al = 8.1/27 = 0.3 mol. Atoms = 0.3 x 6.022x1023. In FCC, 4 atoms per unit cell. Number of unit cells = (0.3 x 6.022x1023)/4 = 4.516x1022.
(b)
(i) In NaCl, sizes of Na+ and Cl- are comparable, so Schottky defect.
(ii) Phosphorus has 5 valence electrons; extra electron is free, creating n-type semiconductor.
(iii) In ferrimagnetic substances, unequal magnetic moments don't cancel; in antiferromagnetic, equal and opposite moments cancel out.
2014 · 5 marks
(a) Define the following terms:
(i) Molarity
(ii) Molal elevation constant (Kb)
(b) A solution containing 15 g of urea (molar mass = 60 g mol−1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose (molar mass = 190 g mol−1) in water. Calculate the mass of glucose present in one litre of its solution.
Answer(a)
(i) Molarity: Number of moles of solute present in one litre of the solution.
(ii) Molal elevation constant (Kb): Elevation of the boiling point of a solution when one mole of a non-volatile solute is dissolved in one kilogram of a volatile solvent.
(b) πurea = πglucose
Curea = Cglucose nurea/V = nglucose/V 15/60 = Wglucose/190 Wglucose = 15 × 190/60 = 47.5 g (Note: if molar mass of glucose = 180, then W = 45 g)
2014 · 5 marks
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol−1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution? (Density of solution = 1.2 g mL−1)
Answer(a) Positive deviation from Raoult's law. The intermolecular forces A–B < A–A and B–B. Here A is solute and B is solvent, so A–B shows weaker interaction than the pure components.
(b) 10% by mass means 10 g glucose in 100 g solution.
Moles of glucose = 10/180 = 0.056 mol Mass of solvent = 90 g = 0.09 kg Molality = 0.056/0.09 = 0.62 m Volume of solution = 100/1.2 = 83.33 mL = 83.33 × 10−3 L Molarity = 0.056/(83.33 × 10−3) = 0.67 M
2016 · 5 marks
(a) Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g/mol) was dissolved in 50 g of water, assuming complete ionization. (Kf for water = 1.86 K kg/mol). (b)
(i) Out of 1M glucose and 2M glucose, which has higher boiling point and why?
(ii) What happens when external pressure exceeds osmotic pressure of solution?
OR
(a) When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (Sx). (Kf for CS2 = 3.83 K kg/mol, Atomic mass of S = 32 g/mol).
(b) Blood cells are isotonic with 0.9% NaCl. What happens in:
(i) 1.2% NaCl?
(ii) 0.4% NaCl?
Answer(a) MgCl2 -> Mg2+ + 2Cl-, i = 3. m = (1.9 x 1000)/(95 x 50) = 0.4 mol/kg. delta-Tf = 3 x 1.86 x 0.4 = 2.232 K. Tf = 273.15 - 2.232 = 270.918 K. (b)
(i) 2M glucose has higher boiling point due to greater elevation.
(ii) Reverse osmosis takes place.
OR
(a) MB = (3.83 x 2.56 x 1000)/(0.383 x 100) = 256 g/mol. n = 256/32 = 8. Formula: S8. (b)
(i) Cells shrink (crenation) - hypertonic.
(ii) Cells swell (haemolysis) - hypotonic.
2017 · 5 marks
(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.
Given: (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1)
(b) Define the following terms:
(i) Molality (m)
(ii) Abnormal molar mass
OR
(a) 30 g of urea (M = 60 g mol-1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.
(b) Write two differences between ideal solutions and non-ideal solutions.
(i) Why is boiling point of 1M NaCl solution more than that of 1M glucose solution?
(ii) A nonvolatile solute 'X' (molar mass = 50 g mol-1) when dissolved in 78g of benzene reduced its vapour pressure to 90%. Calculate the mass of X dissolved in the solution.
(iii) Calculate the boiling point elevation for a solution prepared by adding 10g of MgCl2 to 200g of water assuming MgCl2 is completely dissociated. (Kb for Water = 0.512 K kg mol-1, Molar mass MgCl2 = 95g mol-1)
Answer(a)
(i) NaCl is having ionic bonding between sodium and chloride which is strong bonding while glucose having covalent bonding which is weak in nature. NaCl will require high temperature to boil while glucose needs low temperature to dissociate its bonding. NaCl dissociates into 2 ions (Na+ and Cl-) giving van't Hoff factor i=2, while glucose does not dissociate (i=1).
(ii) (P°-Ps)/P° = n/N = (w x M)/(m x w_solvent)
(100-90)/90 = (w x 78)/(50 x 78) 10/90 = w/50 w = (10 x 50)/90 = 5.55 grams
(iii) Tb = i x Kb x m (MgCl2 = 3 ions, i = 3)
Tb = (3 x 0.512 x 10)/(95 x 0.2) = 5.36/19 = 0.80 So elevation in boiling point = 273 + 0.80 = 273.80 K
2023 · 5 marks
33 (OR): (b)
(i) Why is the value of Van't Hoff factor for ethanoic acid in benzene close to 0.5?
(ii) Determine the osmotic pressure of a solution prepared by dissolving 2.32 x 10-2g of K2SO4 in 2L of solution at 25°C, assuming that K2SO4 is completely dissociated. (R = 0.082 L atm K-1 mol-1, Molar mass K2SO4 = 174g mol-1)
(iii) When 25.6g of sulphur was dissolved in 1000g of benzene, the freezing point lowered by 0.512 K. Calculate the formula of sulphur (Sx). (Kf for benzene = 5.12 K kg mol-1, Atomic mass of Sulphur = 32g mol-1)
Answer(b)
(i) Van't Hoff factor is used in determination of colligative property. It is also called Abnormal colligative property. i = actual no. of particles / theoretical number of particles. It is considered that degree of dissociation of ethanoic acid in benzene is 1. 2CH3-COOH -> (CH3-COOH)2. So, i = 1/2 = 0.5
(ii) K2SO4 -> 2K+ + SO42-, ions produced = 3, i = 3
pi = iCRT = 3 x (0.0232/174) x 0.0821 x 298 = 0.851/174 = 0.00489 = 4.89 x 10-3 atm
(iii) Delta Tf = (Kf x m x 1000)/(MB x WA) where MB = Sx
0.512 = (5.12 x 25.6 x 1000)/(Sx x 1000) Sx = (5.12 x 25.6)/0.512 = 256 x = 256/32 = 8 So the formula of sulphur is S8.
2024 · 5 marks
(a)
(i) Ishan's automobile radiator is filled with 1.0 kg of water. How many grams of ethylene glycol (Molar mass = 62 g mol-1) must Ishan add to get the freezing point of the solution lowered to -2.8°C. Kf for water is 1.86 K kg mol-1.
(ii) What type of deviation from Raoult's law is shown by ethanol and acetone mixture? Give reason.
OR (b)
(i) Boiling point of water at 750 mm Hg pressure is 99.68°C. How much sucrose (Molar mass = 342 g mol-1) is to be added to 500 g of water such that it boils at 100°C? (Kb for water = 0.52 K kg mol-1)
(ii) State Henry's law and write its any one application.
Answer(a)
(i) Delta_Tf = Kf x m. 2.8 = 1.86 x m/(62 x 1). m = 2.8 x 62/1.86 = 93.3 g. Mass of ethylene glycol required = 93.3 g.
(ii) Positive Deviation. When a solution shows positive deviation, its vapor pressure is greater than what Raoult's law predicts. Pure ethanol possesses hydrogen bonding interactions. When acetone is introduced into the mixture, it disrupts some of the hydrogen bonds between ethanol molecules, leading to positive deviation.
OR (b)
(i) Delta_Tb = Kb x m. 0.32 = 0.52 x W x 1000/(342 x 500). W = 0.32 x 342 x 500/(0.52 x 1000) = 105.23 g. Weight of sucrose to be added = 105.23 g.
(ii) Henry's law states that the solubility of the gas is directly proportional to the pressure of the gas at a constant temperature. Application: In soft drinks, to increase the solubility of CO2, the bottles are sealed at a high pressure.
2024 · 5 marks
(a)
(i) Define reverse osmosis.
(ii) Why are aquatic species more comfortable in cold water in comparison to warm water?
(iii) A solution containing 2 g of glucose (M = 180 g mol-1) in 100 g of water is prepared at 303 K. If the vapour pressure of pure water at 303 K is 32.8 mm Hg, what would be the vapour pressure of the solution?
OR (b)
(i) Predict whether Van't Hoff factor will be less or greater than one, when ethanoic acid is dissolved in benzene.
(ii) Define ideal solution.
(iii) Calculate the mass of CaCl2 (molar mass = 111 g mol-1) to be dissolved in 500 g of water to lower its freezing point by 2K, assuming that CaCl2 undergoes complete dissociation. (Kf for water = 1.86 K kg mol-1)
Answer(a)
(i) Reverse osmosis is the process of moving a solvent from a solution to a pure solvent via a semipermeable membrane while applying excess pressure on the solution side.
(ii) In colder water, the solubility of oxygen increases. Aquatic species with higher oxygen demands find colder water more comfortable because it provides a greater supply of dissolved oxygen.
(iii) Number of moles of water = 100/18 = 5.55 moles. Number of moles of glucose = 2/180 = 0.0111. Total moles = 5.56. X_water = 5.55/5.56. P_solution = X_water x P°_water = 0.99 x 32.8 = 32.74 mm of Hg.
OR (b)
(i) Less than one. In benzene, ethanoic acid molecules associate through hydrogen bonding rather than dissociate into ions. Its Van't Hoff factor is less than 1.
(ii) It is defined as the solution that obeys Raoult's Law, there is no interaction between the molecules, no volume change or heat change upon mixing.
(iii) Delta_Tf = i x Kf x molality. CaCl2 -> Ca2+ + 2Cl-, i = 3. 2 = 3 x 1.86 x (w/111) x (1000/500). w = 2 x 111 x 500 / (3 x 1.86 x 1000) = 19.89 g.