Question

The solubility product of a sparingly soluble $A B_{2}$ salt is $2.56 \times 10^{-4} M^{3}$ at $25^{\circ} C$. The $K_{f}$ of water is $1.8\, kg\, mol ^{-}$. The depression in freezing point of a standard solution of $A B_{2}$ is

• A. 0.432 K
• B. 0.216 K
• C. 0.108 K
• D. 13.824 K

Answer 1

Answer: (B)

Let, solubility of $A B_{2}(1: 2$ type electrolyte $)$ is pure water $=5\, mol\, L^{-1}=5\, M$
$\Rightarrow K_{s p}=4 S^{3}=2.56 \times 10^{-4} M^{3}$ (given)
$\therefore S=0.04\, M=0.04\, m$ (molal) under standard condition of the solution.
Depression of freezing point,
$\Delta T_{f}=K_{f} \times m \times i=1.8 \times 0.04 \times 3$
$=0.216\, K$
[ $\because$ Assuming complete dissociation of $A B_{2}$ van't Hoff factor, $i=3$]