Q16 / 90 · JEE Main 2019 (12 Jan Shift 2) · Chemistry

$8 g$ 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:

Correct answer
A
$0.167, 11.11$
B
$0.167, 22.20$
C
$0.2, 11.11$
D
$0.2, 22.20$

Solution

n_{1} = \frac{8}{40} = 0.2

n_{2} = \frac{18}{18} = 1

;

n_{t} = n_{1} + n_{2} = 0.2 + 1 = 1.2

Mole fraction of

N a O H = \frac{n_{1}}{n_{t}} = \frac{0.2}{1.2} = 0.167

Molality

= \frac{8}{40} \times \frac{1000}{18} = 11.11

Hence, the correct answer is option: 'A'.

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