Some Basic Concepts of Chemistry 1 Question 29
11. A solution of sodium sulphate contains $92 \mathrm{~g}$ of $\mathrm{Na}^{+}$ions per kilogram of water. The molality of $\mathrm{Na}^{+}$ions in that solution in $\mathrm{mol} \mathrm{kg}^{-1}$ is
(2019 Main, 9 Jan I)
(a) 16
(b) 4
(c) 132
(d) 8
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Solution:
- Molality $(m)=\frac{\text { Number of moles of solute }}{\text { Mass of solvent (in g) }} \times 1000$
$$ \begin{aligned} & =\frac{\text { Mass of solute }(\text { in } \mathrm{g}) \times 1000}{\left[\begin{array}{r} \text { Molecular weight of solute } \ \times \text { mass of solvent }(\text { in } \mathrm{g}) \end{array}\right]} \ & =\frac{w_{\mathrm{Na}^{+}} \times 1000}{M_{\mathrm{Na}^{+}} \times w_{\mathrm{H}_{2} \mathrm{O}}}=\frac{92 \times 1000}{23 \times 1000}=4 \mathrm{~mol} \mathrm{~kg}^{-1} \end{aligned} $$
