Question

$0.48\, g$ of a substance is dissolved in $10.6\, g$ of $C_{6}H_{6}$. The freezing point of benzene is lowered by $1.8^{\circ}C$ What will be the mol. wt. of the substance $(k_{f}$ for benzene = 5)

• A. $250.2$
• B. $90.8$
• C. $125.79$
• D. $102.5$

Answer 1

Answer: (C)

$\Delta T_{f} =\frac{1000\,K_{f} w}{mw}$
$m=\frac{1000\,k_{f} w}{\Delta T_{f} w}$
$=\frac{1000\times5\times0.48}{1.8\times10.6}$
$=125.79$