Physical Chemistry XII · Chapter Practice

Chapter Practice

MCQ practice for Solutions, Electrochemistry & Chemical Kinetics — click an option to check your answer.

💧 Solutions 7 marks

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Which of the following is a colligative property?

Correct: C — Osmotic pressure. Colligative properties depend only on the number of solute particles, not their nature. The four colligative properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Molality is defined as:

Correct: B. Molality (m) = moles of solute / mass of solvent in kg. Unlike molarity, molality does not change with temperature (since it uses mass, not volume). This makes it preferred for colligative property calculations.

An azeotropic mixture of HNO₃ (68%) and water boils at 393.5 K. This is a ______ deviation from Raoult's Law.

Correct: B. HNO₃–H₂O shows negative deviation (A–B interactions stronger than A–A and B–B) → lower vapour pressure → higher boiling point → maximum boiling azeotrope. Compare: ethanol–water shows positive deviation → minimum boiling azeotrope (bp 351.3K).

If Kb for water = 0.52 K kg mol⁻¹, the boiling point elevation for 0.5 molal glucose solution is:

Correct: A — 0.26 K. ΔTb = Kb × m = 0.52 × 0.5 = 0.26 K. Glucose is a non-electrolyte, so van't Hoff factor i = 1. No correction needed.

The van't Hoff factor i for MgCl₂ that is 80% dissociated is:

Correct: B — 2.6. MgCl₂ → Mg²⁺ + 2Cl⁻ (n = 3 ions). i = 1 + (n − 1)α = 1 + (3 − 1) × 0.8 = 1 + 1.6 = 2.6.

Henry's law constant KH increases with temperature. This implies that solubility of a gas in liquid:

Correct: B. Henry's law: p = KH × x (mole fraction). If KH increases at higher T, then for the same partial pressure, the mole fraction (solubility) must decrease. Gases become less soluble in liquids as temperature rises.

Which of the following is a non-ideal solution showing negative deviation from Raoult's Law?

Correct: A — CHCl₃ + acetone. Chloroform and acetone form H-bonds with each other (A–B > A–A and B–B) → negative deviation. Benzene + toluene is nearly ideal. Acetone + ethanol and ethanol + cyclohexane show positive deviation.

Reverse osmosis is used in water purification. Which condition is applied?

Correct: C. In reverse osmosis, external pressure greater than the osmotic pressure is applied on the solution side, forcing solvent molecules to pass through the semipermeable membrane from solution to pure solvent — i.e., opposite to normal osmosis.

⚡ Electrochemistry 9 marks

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E°cell for the Zn–Cu cell (E°(Cu²⁺/Cu) = +0.34V, E°(Zn²⁺/Zn) = −0.76V) is:

Correct: B — +1.10 V. E°cell = E°cathode − E°anode = (+0.34) − (−0.76) = +1.10 V. Zn is oxidised (anode), Cu²⁺ is reduced (cathode). A positive E°cell means the reaction is spontaneous.

During electrolysis of aqueous NaCl, the product at the cathode is:

Correct: C — H₂ gas. In aqueous NaCl, at cathode: 2H₂O + 2e⁻ → H₂ + 2OH⁻ (water is reduced in preference to Na⁺, as Na reduction requires very high potential). At anode: 2Cl⁻ → Cl₂ + 2e⁻.

96500 C of charge deposits 108 g of silver (M = 108 g/mol). To deposit 54 g, the required charge is:

Correct: B — 48250 C. By Faraday's first law: mass deposited ∝ charge. To deposit half the mass (54g instead of 108g), we need half the charge: 96500/2 = 48250 C. (Ag⁺ + e⁻ → Ag, so 1 Faraday deposits 1 mol = 108g.)

Kohlrausch's law states that at infinite dilution:

Correct: B. Kohlrausch's law: Λ°m = ν₊λ°₊ + ν₋λ°₋. At infinite dilution, ions are independent. This is used to calculate Λ°m for weak electrolytes indirectly, since they cannot be extrapolated to c = 0 from Λm vs √c graphs.

The Nernst equation for a cell at 298K is written as:

Correct: B. E = E° − (RT/nF)ln Q = E° − (0.0591/n) log Q at 298K. As the cell discharges, Q increases, so E decreases. At equilibrium, E = 0 and Q = K, giving log K = nE°/0.0591.

Molar conductivity of weak electrolytes ______ with dilution:

Correct: C. Weak electrolytes (like CH₃COOH) are partially dissociated. On dilution, degree of dissociation increases sharply, so molar conductivity increases steeply near infinite dilution. Strong electrolytes increase gradually (√c relationship).

The unit of cell constant is:

Correct: A — cm⁻¹. Cell constant G* = l/A (distance between electrodes / area of cross-section). Conductance G = κ/G*, so G* = κ/G. Units: (S cm⁻¹)/(S) = cm⁻¹. It is a property of the cell geometry, not the solution.

Which of the following correctly relates ΔG° to E°cell?

Correct: D. ΔG° = −nFE°cell. For a spontaneous reaction, E°cell > 0 and ΔG° < 0 (consistent). Also, ΔG° = −RT ln K, so E°cell = (RT/nF) ln K = (0.0591/n) log K at 298K.

In a fuel cell, H₂ and O₂ react to produce electricity. The product at the cathode is:

Correct: C — Water. In H₂–O₂ fuel cell: Anode: H₂ → 2H⁺ + 2e⁻. Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O. Overall: 2H₂ + O₂ → 2H₂O. Fuel cells are highly efficient and produce no harmful by-products.

Q10

Which of the following has the highest molar conductance at infinite dilution (Λ°m)?

Correct: B — HCl. H⁺ has exceptionally high ionic conductance (λ° = 350 S cm² mol⁻¹) due to the Grotthuss mechanism (proton hopping through water). OH⁻ is second highest. HCl has Λ°m ≈ 426 S cm² mol⁻¹, far higher than NaCl (126) or KCl (150).

⏱ Chemical Kinetics 7 marks

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For a first-order reaction, the half-life:

Correct: B. For first order: t½ = 0.693/k. It depends only on k (rate constant), not on initial concentration. For zero order: t½ = [A]₀/2k (depends on [A]₀). For second order: t½ = 1/(k[A]₀) (inversely proportional).

The unit of rate constant for a second-order reaction is:

Correct: C — L mol⁻¹ s⁻¹. Units of k = mol^(1-n) L^(n-1) s⁻¹, where n = order. For n=2: mol^(-1) L^1 s⁻¹ = L mol⁻¹ s⁻¹. For n=1: s⁻¹. For n=0: mol L⁻¹ s⁻¹.

The slope of a log k vs 1/T graph (Arrhenius equation) is:

Correct: B — −Ea/2.303R. From Arrhenius: log k = log A − Ea/(2.303RT). This is y = c + mx form, where y = log k, x = 1/T, and slope = −Ea/2.303R. Since slope is negative, the graph slopes downward. Ea = −2.303R × slope.

For a reaction A → B, doubling [A] doubles the rate. The order with respect to A is:

Correct: C — First order. Rate = k[A]^n. If [A] doubles and rate doubles: 2 = 2^n, so n = 1. If rate quadrupled when [A] doubled: 4 = 2^n → n = 2. If rate is unchanged: n = 0 (zero order).

Activation energy of a reaction can be decreased by:

Correct: B — Catalyst. A catalyst provides an alternative reaction pathway with lower activation energy. Increasing temperature increases the fraction of molecules having energy ≥ Ea, but does NOT decrease Ea itself. Concentration affects rate but not Ea.

Which graph gives a straight line for a first-order reaction?

Correct: C — log[A] vs t. Integrated rate law for first order: log[A] = log[A]₀ − kt/2.303. This is linear (y = c − mx). For zero order: [A] vs t is linear. For second order: 1/[A] vs t is linear.

For the reaction: 2N₂O₅ → 4NO₂ + O₂, if rate = −½ d[N₂O₅]/dt, then rate in terms of O₂ is:

Correct: B. Rate = −(1/2)d[N₂O₅]/dt = +(1/4)d[NO₂]/dt = +d[O₂]/dt. From stoichiometry, for every 2 moles of N₂O₅ consumed, 1 mole of O₂ is produced. Rate = d[O₂]/dt.

The pre-exponential factor A in the Arrhenius equation represents:

Correct: B. In k = A·e^(−Ea/RT), A is the frequency factor (or pre-exponential factor) that accounts for the frequency of collisions and the proper orientation of reactant molecules. It has the same units as k. At T → ∞, e^(−Ea/RT) → 1, so k → A.