Question

$8.0575\times {10}^{-2}kg$ of Glauber's salt is dissolved in water to obtain $1d{m}^3$ of solution of density $1077.2kg{m}^{-3}$. Calculate the molality, molarity and mole fraction of $N{a}_{2}S{O}_4$ in solution.

Answer 1

Molar mass of Glauber's salt $(N{a}_{2}S{O}_4.10H_{2}O)$
$=23\times 2+32+64+10\times 18=322g$
$\Rightarrow$ Mole of $N{a}_{2}S{O}_4.10H_{2}O$ in $1.0L$ solution $=\frac{80.575}{322}=0.25$
$\Rightarrow$ Molarity of solution $=0.25M$
Also, weight of $1.0L$ solution $=1077.2g$
weight of $N{a}_{2}S{O}_4$ in $1.0L$ solution $=0.25\times 142=35.5g$
$\Rightarrow$ Weight of water in $1.0L$ solution $=1077.2-35.5=1041.7g$
$\Rightarrow$ Molality $=\frac{0.25}{1041.7}\times 1000=0.24m$
Mole fraction of $N{a}_{2}S{O}_4=\frac{\text{Mole of }N{a}_{2}S{O}_4}{\text{Mole of }N{a}_{2}S{O}_4+\text{ Mole of water}}$
$=\frac{0.25}{0.25+\frac{1041.7}{18}}=4.3\times 10^{-3}$