# $8.0575 × 10^{–2}$ kg of Glauber’s salt ($Na_2SO_4.10H_2O$) is dissolved in water to obtain 1 $dm^3$ of a solution of density $1077.2 kgm^{–3}$. Calculate the mole fraction of $Na_2SO_4$ in the solution.

$X_1=\large\frac{n_1}{n_1+n_2}$
$n_1=\large\frac{80.575}{322}$$=0.25 Weight of solution =\large\frac{1000\times 1077.2\times 1000}{10^6}=$$1077.2g$
Weight of solvent =1077.2-80.575
$\Rightarrow 996.625$
Moles of water=$\large\frac{996.625}{18}$$=55.3 X_1=\large\frac{0.25}{0.25+55.3}=\frac{0.25}{55.6}$$=0.0043$
Hence (A) is the correct answer.