Question

Van't Hoff factor, i, of a 0.5% (w/w) aqueous solution of KCl which freezes at $-{0.24}^{o}C$ is:

$[K_f$ of water = 1.86 K kg mol${}^{-1}$, Mol.wt.of KCl = 74.5]

• A. 1.52
• B. 2.32
• C. 1.92
• D. 1.32

Answer 1

Answer: (C)

1.92

$\Delta T_f$ = $K_f\times m$

$m=1000\times \frac{weightofsolute}{(mol.wt.ofKCl\times weightofsolvent)}=1000\times \frac{0.5}{(74.5\times 99.5)}=0.067m$

$\Delta T_f=1.86\times 0.067=0.125$ ( calculated value )

$\Delta T_f$ = ($T_{f_1}-T_{f_2})=0-(-0.24)=0.24$ ( observed value )

Vant Hoff factor , $i=\frac{observedvalueof\Delta T_f}{calculatedvalueof\Delta T_f}=\frac{0.24}{0.125}=1.92$

Hence, the correct option is $\text{C}$