Some Basic Concepts of Chemistry 1 Question 26
47. $8.0575 \times 10^{-2} \mathrm{~kg}$ of Glauber’s salt is dissolved in water to obtain $1 \mathrm{dm}^{3}$ of solution of density $1077.2 \mathrm{~kg} \mathrm{~m}^{-3}$. Calculate the molality, molarity and mole fraction of $\mathrm{Na}{2} \mathrm{SO}{4}$ in solution.
(1994, 3M)
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Solution:
- Molar mass of Glauber’s salt $\left(\mathrm{Na}{2} \mathrm{SO}{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}\right)$
$$ =23 \times 2+32+64+10 \times 18=322 \mathrm{~g} $$
$\Rightarrow$ Mole of $\mathrm{Na}{2} \mathrm{SO}{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}$ in $1.0 \mathrm{~L}$ solution $=\frac{80.575}{322}=0.25$
$\Rightarrow$ Molarity of solution $=0.25 \mathrm{M}$
Also, weight of $1.0 \mathrm{~L}$ solution $=1077.2 \mathrm{~g}$
weight of $\mathrm{Na}{2} \mathrm{SO}{4}$ in $1.0 \mathrm{~L}$ solution $=0.25 \times 142=35.5 \mathrm{~g}$
$\Rightarrow$ Weight of water in $1.0 \mathrm{~L}$ solution $=1077.2-35.5=1041.7 \mathrm{~g}$
$\Rightarrow$ Molality $=\frac{0.25}{1041.7} \times 1000=0.24 \mathrm{~m}$
Mole fraction of $\mathrm{Na}{2} \mathrm{SO}{4}=\frac{\text { Mole of } \mathrm{Na}{2} \mathrm{SO}{4}}{\text { Mole of } \mathrm{Na}{2} \mathrm{SO}{4}+\text { Mole of water }}$
$$ \begin{aligned} & =\frac{0.25}{0.25+\frac{1041.7}{18}} \ & =4.3 \times 10^{-3} . \end{aligned} $$
