Question

Two solvents A and B have $K_f$ values 1.86 and 2.72 K $mo{l}^{-1}$ kg respectively. A given amount of substance when dissolved in 500 g of A, it completely dimerizes and when the same mass of the substance is dissolved in 500 g of B, the solute undergoes trimerization. The ratio of observed lowering of freezing points in two cases is:

• A. 1:2
• B. 2:1
• C. 1:3
• D. 1:1

Answer 1

The correct option is D 1:1

The van't Hoff factor $i$ is given by the expression $i=\frac{\text{number of solute particles present in solution (n observed)}}{\text{theoretical number of solute particles due to solution of non electrolyte (n theoretical)}}$
For solvent A, $i=\frac{1}{2}$ as the solute undergoes dimerization.
For the solvent B, $i=\frac{1}{3}$ as the solute undergoes trimerization.
The depression in the freezing point $\Delta T_f$ for the solution is given by the expression $\Delta T_f=i\times m\times K_f$ .
Here, m is the molality and $K_f$ is the molal depression in the freezing point constant.
Since, the same mass of substance is dissolved in same amount of solvents A and B, the molality of the substance in two solvents is same.
For solvent A, $i\times K_f=\frac{1.86}{2}=0.93$
For solvent B, $i\times K_f=\frac{2.72}{3}=0.91$
For both the solvents $\Delta T_f=i\times m\times K_f$ is same.
Hence, the ratio of observed lowering of freezing points in two cases is 1:1.