Two solvents 'A' and 'B' have values 1.86 and , respectively. A given amount of a substance, when di — Solutions and Colligative Properties Chemistry Question
Question
Two solvents 'A' and 'B' have $K_f$ values 1.86 and $2.79 \text{ K mol}^{-1}\text{ kg}$, respectively. A given amount of a substance, when dissolved in 500 g of 'A', completely dimerizes and when the same amount of the substance is dissolved in 500 g of 'B', the solute undergoes trimerization. The ratio of observed lowering of freezing point in two cases is
💡 Solution & Explanation
Let the amount of substance be $n$ moles. Molality in both cases is $m = n/0.5 = 2n$. In solvent A, it completely dimerizes ($i = 1/2$), so $\Delta T_{f(A)} = i \times K_{f(A)} \times m = (1/2) \times 1.86 \times 2n = 1.86n$. In solvent B, it completely trimerizes ($i = 1/3$), so $\Delta T_{f(B)} = i \times K_{f(B)} \times m = (1/3) \times 2.79 \times 2n = 0.93 \times 2n = 1.86n$. The ratio $\Delta T_{f(A)} : \Delta T_{f(B)}$ is $1.86n : 1.86n = 1:1$. Therefore, correct answer is C.