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. Density of this solution at $298\, K$ is $2.0 \,g cm ^{-3}$. The ratio of the molecular weights of the solute and solvent $\left(\frac{ MW _{\text {solute }}}{ MW _{\text {solvent }}}\right)$, is

Answer 1

$x _{ B }=0.1 ; m = M ; d =2 \,g / cm ^{3} ; \frac{ M _{ B }}{ M _{ A }}=?$
$m =\frac{0.1}{0.9} \times \frac{1000}{ M _{ A }} $
$[ A \rightarrow$ Solvent, $B \rightarrow$ Solute $]$
$d =\frac{ M }{ m }+\frac{ MM _{ B }}{1000}$
$2=1+\frac{1}{9} \times \frac{1000}{ M _{ A }} \cdot \frac{ M _{ B }}{1000}$
$1=\frac{1}{9} \frac{ M _{ B }}{ M _{ A }}$
$\frac{ M _{ B }}{ M _{ A }}=9$