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 $]$
$d=\frac{M}{m}+\frac{M M_{B}}{1000}$
$2=1+\frac{1}{9} \times \frac{1000}{M_{A}} \cdot \frac{M_{B}}{1000}$
$1=\frac{1}{9} \frac{M_{B}}{M_{A}}$
$\frac{M_{B}}{M_{A}}=9$