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JEE Main 201912 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry

JEE Main 2019 · 12 January · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

8g of NaOH is dissolved in 18g of H _ 2 O . Mole fraction of NaOH...

8g

of

NaOH

is dissolved in

18g

of

{H}_{2}O

. Mole fraction of

NaOH

in solution and molality (in

mol{kg}^{- 1}

) of the solution respectively are:

Options

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

Correct answer

A. $0.167, 11.11$

Step-by-step solution

{n}_{1} = \dfrac{8}{40} = 0.2

{n}_{2} = \dfrac{18}{18} = 1

;

{n}_{t} = {n}_{1} + {n}_{2} = 0.2 + 1 = 1.2

Mole fraction of

NaOH = \dfrac{{n}_{1}}{{n}_{t}} = \dfrac{0.2}{1.2} = 0.167

Molality

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

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