Question

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

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

Answer 1

The correct option is C $80$%

Depression in freezing point is given by the formula :

$\Delta T_f=i{K}_f.m$

Molality of $P=\frac{1.25}{94}\times \frac{1000}{50}=\frac{25}{94}m$

$i=\frac{\Delta T_f}{K_f.m}=\frac{0.3\times 94}{1.86\times 25}=0.60$

$2P\rightleftharpoons P_2$

Before dissociation $1$ $0$

After dissociation $1-\alpha$ $\frac{\alpha }{2}$

$\alpha =$ Degree of dissociation

$i=1-\alpha +\frac{\alpha }{2}$

$i=1-\frac{\alpha }{2}$

$0.6=1-\frac{\alpha }{2}$

$\alpha =0.8$

Therefore the degree of dissociation is 80%.

Hence, option C is correct.