Question

A solution of $1.25\, g$ of $P$ in $50 \,g$ of water lowers freezing point by $0.3\,{}^\circ\,C$. Molar mass of $P$ is $94.K_{f(water)}$ = $1.86\, K\, kg \,mol^{-1}$. The degree of association of $'P'$ in water is

• A. $60 \%$
• B. $75 \%$
• C. $80 \%$
• D. $65 \%$

Answer 1

Answer: (C)

Step I Calculation of van't Hoff factor of ' $P'$

$\Delta T_{f}=i \times K_{f} \times m=i \times K_{f} \times \frac{w_{1} \times 1000}{W_{2} \times M_{1}} $

$0.3=i \times 1.86 \times \frac{1.25 \times 1000}{50 \times 94}$

$\Rightarrow i=\frac{0.3 \times 50 \times 94}{1.86 \times 1.25 \times 1000}=0.6064$

Step II Calculation of degree of association of ' $P'$

Degree of association,

$\alpha=\frac{i-1}{\frac{1}{n}-1}=\frac{0.6064-1}{\frac{1}{2}-1}=\frac{-0.3936}{-0.5}=0.7872=78.72 \% \approx 80 \%$