Some Basic Concepts of Chemistry 1 Question 7
7. $8 \mathrm{~g}$ of $\mathrm{NaOH}$ is dissolved in $18 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$. Mole fraction of $\mathrm{NaOH}$ in solution and molality (in $\mathrm{mol} \mathrm{kg}^{-1}$ ) of the solution respectively are
(2019 Main, 12 Jan II)
(a) $0.2,11.11$
(b) $0.167,22.20$
(c) $0.2,22.20$
(d) $0.167,11.11$
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Solution:
- Mole fraction of solute $=\underline{\text { number of moles of solute }+ \text { number of moles solvent }}$ number of moles of solute
$$ \chi_{\text {Solute }}=\frac{n_{\text {Solute }}}{n_{\text {Solute }}+n_{\text {Solvent }}}=\frac{\frac{w_{\text {Solute }}}{M w_{\text {Solute }}}}{\frac{w_{\text {Solute }}}{M w_{\text {Solute }}}+\frac{w_{\text {Solvent }}}{M w_{\text {Solvent }}}} $$
Given, $\quad w_{\text {Solute }}=w_{\mathrm{NaOH}}=8 \mathrm{~g}$
$M w_{\text {Solute }}=M w_{\mathrm{NaOH}}=40 \mathrm{~g} \mathrm{~mol}^{-1}$
$w_{\text {Solvent }}=w_{\mathrm{H}_{2} \mathrm{O}}=18 \mathrm{~g}$
$$ M w_{\text {Solvent }}=18 \mathrm{~g} \mathrm{~mol}^{-1} $$
$\therefore \quad \chi_{\text {Solute }}=\chi_{\mathrm{NaOH}}=\frac{8 / 40}{\frac{8}{40}+\frac{18}{18}}=\frac{0.2}{0.2+1}=\frac{0.2}{1.2}=0.167$
Now, molality $(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent (in } \mathrm{kg})}$
$$ \begin{aligned} & =\frac{\frac{w_{\text {Solute }}}{M w_{\text {Solute }}}}{w_{\text {Solvent }}(\text { in } \mathrm{g})} \times 1000=\frac{\frac{8}{40}}{18} \times 1000 \ & =\frac{0.2}{18} \times 1000=11.11 \mathrm{~mol} \mathrm{~kg}^{-1} \end{aligned} $$
Thus, mole fraction of $\mathrm{NaOH}$ in solution and molality of the solution respectively are 0.167 and $11.11 \mathrm{~mol} \mathrm{~kg}^{-1}$.
