If the boiling points of two solvents X and Y (having same molecular weights) are in the ratio 2: 1 and their enthalpy of vaporizations are in the ratio 1: 2, then the boiling point elevation constant of X is m times the boiling point elevation constant of Y. The value of m is ____________ (nearest integer)

Solution

Chemistry ma’am · 4 steps

Write K_b in terms of T_b and Δ Hᵥₐₚ, then take the ratio. Everything else cancels.

1. Formula for Kb. The boiling point elevation constant depends on the solvent's boiling point, molar mass and enthalpy of vaporisation. Here M is in g mol⁻¹ and Δ Hᵥₐₚ is per mole.

   K_b = (R T_b² M)/(1000 Δ Hᵥₐₚ)

2. Cancel common terms. X and Y have the same molar mass, and R and 1000 are constants. So K_b ∝ T_b²/(Δ Hᵥₐₚ).
3. Put in ratios. K_b(X)/K_b(Y) = (T_X/T_Y)² × (Δ H_Y)/(Δ H_X) = (2)² × 2/1 = 4 × 2.
4. So. K_b of X is 8 times K_b of Y, so m = 8.

The common slip: 4. Forgetting to square the boiling point ratio gives 2 × 2 = 4, not 8.

The idea to keep. K_b grows with the square of the boiling point and falls as Δ Hᵥₐₚ rises. For ratio questions, keep only what differs between the solvents and remember the square on T_b.

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