Question

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

Answer 1

Molar mass of Glauber's salt $ (Na_2SO_4. 10 H_2O)$
$ = 23 \times 2 + 32+ 64 + 10 \times 18 = 322\,g$
$\Rightarrow $ Mole of $ Na_2SO_4. 10\,H_2O $ in $1.0 \, L $ solution $ = \frac{80.575}{322} = 0.25 $
$\Rightarrow $ Molarity of solution $= 0.25 \,M$
Also, weight of $1.0\, L$ solution $= 1077.2\, g$
weight of $Na_2SO_4$ in $1.0\, L$ solution $= 0.25 \times 142 = 35.5 \,g $
$\Rightarrow $ Weight of water in $1.0 \,L$ solution $=1077.2-35.5 = 1041.7 \,g$
$\Rightarrow$ Molality $= \frac{0.25}{1041.7} \times 1000 = 0.24\, m $
Mole fraction of $ Na_2SO_4 = \frac{\text{Mole of } Na_2S O_4 }{\text{Mole of } Na_2SO_4 + \text{ Mole of water}}$
$ = \frac{0.25}{ 0.25 + \frac{1041.7}{18}} =4.3 \times 10^{-3}$