Question

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

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

Answer 1

Answer: (C)

Step I Calculation of van't Hoff factor of ' $P^{\prime }$

$\Delta T_f=i\times K_f\times m=i\times K_f\times \frac{w_1\times 1000}{W_2\times M_1}$

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

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

Step II Calculation of degree of association of ' $P^{\prime }$

Degree of association,

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