Question

An element $X$ (Atomic mass $= 25$) exists as $X_{4}$ is benzene. $51g$ of saturated solution of $X$ in benzene was added to $50.0 g$ of pure benzene. The resulting solution showed a depression of freezing point of $0.55 K$. Find the solubility of $X$ per $100 g$ of benzene. ($K_{f}$ for benzene $= 5.5 \,K\, kg\, mol^{-1}$)

Answer 1

Let $x\, g$ be the mass of element in $51.0\, g$ of saturated solution.
Mass of benzene in $51.0 g$ of saturated solution$ = 51.0 - x g$
Total mass of benzene containing $x g$ of solute$ = 50 + 51 - x = \left(101 - x\right) g$
$\Delta T_{f} = \frac{1000K_{f} W_{B}}{M_{B} W_{A}} = \frac{1000 \times5.5\times x}{4 \times25 \times\left(101 -x\right)}$
$= 0.55$ (given)
$\Rightarrow x = 1.0 g$
Hence, solubility
$= \frac{W_{B} \times100}{W_{A}} = \frac{1}{\left(51 -1\right)} \times100 = 2.0 \,g$