Question

The mole fraction of a solute in a solution is $0.1$ . At $298\,K$ , molarity of this solution is the same as its molality. The density of this solution at $298\,K$ is $2.0\,g\,cm^{- 3}$ . The ratio of the molecular weights of the solute and solvent, $\left(\frac{M W_{s o l u t e}}{M W_{s o l v e n t}}\right)$ , is

Answer 1

$\frac{X_{\text {solute }}}{X_{\text {solvent }}}=\frac{0.1}{0.9}=\frac{1}{9}$
$\frac{W_{\text {solute }}}{W_{\text {solvent }}} \times \frac{M_{\text {solvent }}}{M_{\text {solute }}}=\frac{1}{9} \ldots( i )$
$W_{\text {solute }}+W_{\text {solvent }}=W_{\text {solution }}=$ density $\times$ volume
$W_{\text {solute }}+W_{\text {solvent }}=2 \times V \ldots( ii )$
Molarity $=$ molality
$\frac{n_{\text {solute }}}{V_{\text {solution }}}=\frac{n_{\text {solute }}}{W_{\text {solvent }}}$
$W_{\text {solvent }}=V_{\text {solution }}=\frac{W_{\text {solute }}+W_{\text {solvent }}}{2}$
$2 W_{\text {solvent }}=W_{\text {solute }}+W_{\text {solvent }}$
$W_{\text {solute }}=W_{\text {solvent }} \ldots$ (iii)
Using equation (i) and (iii), we get
$\frac{M_{\text {solute }}}{M_{\text {solvent }}}=9$