(C) Moles of $NaOH = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol$. Moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$. Mole fraction of $NaO

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(C) Moles of $NaOH = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol$. Moles of $H_2O = \frac{18 \ g}{18 \ g \ mol^{-1}} = 1 \ mol$. Mole fraction of $NaOH = \frac{n_{NaOH}}{n_{NaOH} + n_{H_2O}} = \frac{0.2}{0.2 + 1} = \frac{0.2}{1.2} = 0.167$. Molality $(m) = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.2 \ mol}{18 \times 10^{-3} \ kg} = \frac{0.2}{0.018} \approx 11.11 \ mol \ kg^{-1}$.

Class 12 Chemistry Solutions Method of expressing concentration of solution

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$8 \ g$ of $NaOH$ is dissolved in $18 \ g$ of $H_2O$. Mole fraction of $NaOH$ in solution and molality (in $mol \ kg^{-1}$) of the solution respectively are

JEE MAIN 2019 All JEE MAIN

A
$0.2, \ 22.20$
B
$0.2, \ 11.11$
C
$0.167, \ 11.11$
D
$0.167, \ 22.20$

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$8 \ g$ of $NaOH$ is dissolved in $18 \ g$ of $H_2O$. Mole fraction of $NaOH$ in solution and molality (in $mol \ kg^{-1}$) of the solution respectively are

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