# $8.0575\times 10^{-2}Kg$ of Glauber's salt is dissolved in water to obtain $1dm^3$ of a solution of density 1077.2$Kgm^{-3}$.Calculate the molarity ,molality and mole fraction of $Na_2SO_4$ in solution.

$\begin{array}{1 1}(a)\;0.1M,0.15m,2.3\times 10^{-4}\\(b)\;0.2M,0.24m,4.3\times 10^{-3}\\(c)\;0.325M,0.3m,2.3\times 10^{-3}\\(d)\;0.432M,0.123m,1.3\times 10^{-5}\end{array}$

Glauber's salt is $Na_2SO_4.10H_2O$ having mol.wt=322
$\therefore$ weight of $Na_2SO_4$ in $8.0575\times 10^{-1}Kg$ glauber salt
$\Rightarrow \large\frac{142\times 8.0575\times 10^{-2}}{322}$
$\Rightarrow 3.5533\times 10^{-2}Kg$
Molarity 'M' of $Na_2SO_4=\large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times 1}$
$\Rightarrow 0.2502M$
Molality of $Na_2SO_4=\large\frac{Mole\;of\;Na_2SO_4}{wt\;of\;water\;in\;Kg}$
$\Rightarrow \large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}\times \Large\frac{1041.667}{10^3}}$
$\Rightarrow 0.24m$
Mole fraction of $Na_2SO_4=\large\frac{Mole\;of\;Na_2SO_4}{Mole\;of\;Na_2SO_4+mole\;of\;H_2O}$
$\Rightarrow \large\frac{\Large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}}}{\Large\frac{3.5533\times 10^{-2}}{142\times 10^{-3}}+\frac{1041.667}{18}}$
$\Rightarrow 4.3\times 10^{-3}$
Hence (b) is the correct answer.