Some Basic Concepts of Chemistry - Result Question 7
####7. $8 g$ of $NaOH$ is dissolved in $18 g$ of $H _2 O$. Mole fraction of $NaOH$ in solution and molality (in $mol kg^{-1}$ ) 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, $\quad w _{\text {Solute }}=w _{NaOH}=8 g$
$M w _{\text {Solute }}=M w _{NaOH}=40 g mol^{-1}$
$w _{\text {Solvent }}=w _{H _2 O}=18 g$
$$ M w _{\text {Solvent }}=18 g mol^{-1} $$
$\therefore \quad \chi _{\text {Solute }}=\chi _{NaOH}=\frac{8 / 40}{\frac{8}{40}+\frac{18}{18}}=\frac{0.2}{0.2+1}=\frac{0.2}{1.2}=0.167$
Now, molality $(m)=\frac{\text { Moles of solute }}{\text { Mass of solvent (in } kg)}$
$$ \begin{aligned} & =\frac{\frac{w _{\text {Solute }}}{M w _{\text {Solute }}}}{w _{\text {Solvent }}(\text { in } g)} \times 1000=\frac{\frac{8}{40}}{18} \times 1000 \ & =\frac{0.2}{18} \times 1000=11.11 mol kg^{-1} \end{aligned} $$
Thus, mole fraction of $NaOH$ in solution and molality of the solution respectively are 0.167 and $11.11 mol kg^{-1}$.
