Question

The mole fraction of a solute in a solution is $0.1$ . At $298K$ , molarity of this solution is the same as its molality. The density of this solution at $298K$ is $2.0gc{m}^{-3}$ . The ratio of the molecular weights of the solute and solvent, $\left(\frac{M{W}_{solute}}{M{W}_{solvent}}\right)$ , is

Answer 1

Answer: 9

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