Question

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

Answer 1

Answer: 9

$x_B=0.1;m=M;d=2g/c{m}^3;\frac{M_B}{M_A}=?$
$m=\frac{0.1}{0.9}\times \frac{1000}{M_A}$
$[A\to$ Solvent, $B\to$ Solute $]$
$d=\frac{M}{m}+\frac{M{M}_B}{1000}$
$2=1+\frac{1}{9}\times \frac{1000}{M_A}\cdot \frac{M_B}{1000}$
$1=\frac{1}{9}\frac{M_B}{M_A}$
$\frac{M_B}{M_A}=9$