Question

Equimolal solutions A and B show depression in freezing point in the ratio 2:1. A remains in the normal state in solution B will be :

• A. Normal in solution
• B. Dissociated in solution
• C. Associated in solution
• D. Hydrolysed in solution

Answer 1

The correct option is C Associated in solution


Equimolal solutions A and B show depression in freezing point in the ratio 2:1. A remains in the normal state in solution B will be associated in the solution.

The depression in the freezing point $\Delta T_f=i{k}_{f}m$

Since two solutions have the same molality (m), the depression in the freezing point will be proportional to the van't Hoff factor.
$\frac{\Delta T_f\left(A\right)}{\Delta T_f\left(B\right)}=\frac{i_A}{i_B}$.....(1)

A remains in the normal state in solution

$i_A=1$

Equimolal solutions A and B show depression in freezing point in the ratio 2:1.

$\frac{\Delta T_f\left(A\right)}{\Delta T_f\left(B\right)}=\frac{2}{1}$

Substitute these in equation (1).

$\frac{2}{1}=\frac{1}{i_B}$

$i_B=\frac{1}{2}$

Since the van't Hoff factor for solution B is less than 1, it undergoes association.

Hence, the correct option is $C$.