1.22 g of an organic acid is separately dissolved in 100 g of benzene (K b = 2.6 K kg mol⁻¹) and 100 g of acetone (K b = 1.7 K kg mol⁻¹). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by 0.17°C. The increase in boiling point of solution in benzene in °C is x × 10⁻². The value of x is ______________. (Nearest integer) [Atomic mass : C = 12.0, H = 1.0, O =16.0]

Solution

Chemistry ma’am · 4 steps

Get the molar mass from acetone, where nothing pairs up, then halve the particles in benzene.

1. Molar mass from acetone. In acetone the acid stays a monomer, so i = 1. With Δ T_b = K_b m: 0.17 = 1.7 × m, which gives m = 0.1 mol kg⁻¹.
2. Find M. Moles of acid = 0.1 × 0.1 kg = 0.01 mol, so M = 1.22/0.01 = 122 g mol⁻¹. This matches benzoic acid, C₆H₅COOH.
3. Dimer in benzene. In benzene two molecules join into one dimer, so the number of particles halves and i = 1/2. The molality is still 0.1 mol kg⁻¹, because the solute mass and solvent mass are unchanged.

   Δ T_b = i K_b m = 1/2 × 2.6 × 0.1 = 0.13 °C

4. So. Δ T_b = 13 × 10⁻² °C, so x = 13.

The common slip: 26. Forgetting that the acid dimerises in benzene, so using i = 1, gives Δ T_b = 0.26 °C and x = 26 instead of 13.

The idea to keep. When the same solute goes into two solvents, use the solvent where it behaves normally to find the molar mass. Then apply the van't Hoff factor in the other one: a dimer means i = 1/2.

Answer from the NTA final answer key. Solution by PrepWiser.