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