JEE Main 2022 · 26 July · evening · Numerical answer · +4 / −0
The elevation in boiling point for 1 molal solution of non-volatile solute A is 3 K. The depression in freezing point for 2 molal solution of A in the same solvent is 6 K. The ratio of K_b and K_f i.e., K_b / K_f is 1: X. The value of X is [nearest integer]
Answer and solution
Answer 1
Solution
Chemistry ma’am · 4 steps
Divide each temperature change by its own molality to get the constant, then compare the two.
- Two formulas. For a non-volatile, non-electrolyte solute, Δ T_b = K_b m and Δ T_f = K_f m. Each constant is the temperature change divided by the molality.
- Find Kb. The 1 molal solution raises the boiling point by 3 K, so K_b = 3/1 = 3 K kg mol⁻¹.
- Find Kf. The 2 molal solution lowers the freezing point by 6 K, so K_f = 6/2 = 3 K kg mol⁻¹.
K_b/K_f = 3/3 = 1 : 1
- So. The ratio 1 : X is 1 : 1, so X = 1.
The common slip: 2. Comparing the raw changes, 3 K against 6 K, without dividing 6 by the molality 2 gives a ratio of 1 : 2, so X = 2.
The idea to keep. K_b and K_f are per-molal values. Before comparing two colligative effects, always divide each by its own molality, because a bigger change may just come from a more concentrated solution.
Answer from the NTA final answer key. Solution by PrepWiser.