Question

To $500 \,cm ^{3}$ of water, $3 \times 10^{-3} kg$ of $CH _{3} COOH$ is added. If $23 \%$ of $CH _{3} COOH$ is disociated, what will be the depression in freezing point- $\left(K_{f}\right.$ for water is $1.86\, k\, kg$ $mol ^{-1}$ )

• A. 0.372 K
• B. 0.228 K
• C. 0.328 K
• D. 0.556 K

Answer 1

Answer: (B)

Apply $\Delta T _{ f }= i K _{ f } m$

$\Delta T _{ f }= i \times 1.86$

$\Delta T _{ f }= i \times 1.86 \times \frac{\frac{3}{60}}{\frac{500}{1000}} \ldots \ldots(1)$

($\rho$ of $H _{2} O =1 \,g / cm ^{3}$)

So, $500 \,cm ^{3}$ water will weigh $500 \,g$

Also, $\alpha=\frac{i-1}{n-1}$

$\Rightarrow 0.23=\frac{i-1}{2-1}$

$i=1.23$

Placing the value of (i) in equation $1,$ we get

$\Delta T _{ f }=1.23 \times 1.86 \times \frac{3}{60} \times \frac{1000}{500}$

$\Delta T _{ f }=0.228 \,K$