Question

An element $X$ (Atomic mass = 25) exists as $X_4$ in benzene. 51 g of saturated solution of $X$ in benzene was added to 50.0 g of pure benzene. The resulting solution showed a depression in freezing point of 0.55 K. The solubility (in g) of $X$ per 100 g of benzene is:
$(K_f\text{for benzene}=5.5{\text{K kg mol}}^{–1})$
(Assume 100% association takes place.)

Answer 1

Let x be the mass of element in 51.0 g of saturated solution.
Mass of benzene in 51.0 g of saturated solution = $(51-x)$ g
Total mass of benzene containing $x$ g of solute = $(50+51-x)$ g $=(101-x)$ g.
In benzene, it is given that $X$ exist as $X_4$.
$4X\to X_4$
$(\alpha =1\&n=4)$
Van't Hoff factor in case of association is given by,
$i=1-\alpha +\frac{\alpha }{n}$
$=1-1+\frac{1}{4}=\frac{1}{4}$
$△T_f=i.K_f.m=i.K_f.\frac{n_{\text{solute}}}{w_{\text{solvent (kg)}}}$

$0.55=\frac{1}{4}\times 5.5\times \frac{x\times 1000}{25\times (101-x)}$
$x=1$ g
Hence, solubility of $X$ per 100 g of benzene
$=\frac{m_{\text{solute}}\times 100}{m_{\text{solvent}}}=\frac{1\times 100}{(51-1)}=2$ g