The solubility of N _ 2 in water at 300K and 500torr partial pressure is~ 0 .01 gL^ - 1 . The solubility (in g L ^ - 1 ) at 750torr partial pressure is:

Correct answer: D. 0.015 p = K _ H X where p → Partial pressure of gas X _ 2 → Mole fraction of gas in solution K _ H = Henry's law constant So, p ∝ solubility p _ 1 p _ 2 = s _ 1 s _ 2 ⇒ (500)/(0 . 01) = (750)/(x) ∴ x = 0 . 015g / L

JEE Main 20169 AprilOnline

Exam: JEE Main
Year: 2016
Date: 9 April
Mode: Online
Subject: Chemistry
Chapter: Solutions

JEE Main 2016 · 9 April · Online · Chemistry · Solutions

The solubility of N _ 2 in water at 300K and 500torr partial...

The solubility of

{N}_{2}

in water at

\text{300}\text{K}

and

\text{500}\text{torr}

partial pressure is~

0 .01 gL^{- 1} .

The solubility (in

{\text{g L}}^{- \text{1}}

) at

\text{750}\text{torr}

partial pressure is:

Options

A $0.0075$
B $0.005$
C $0 . 02$
D $0.015$

Correct answer

D. $0.015$

Step-by-step solution

p = {K}_{H}X

where

p \to

Partial pressure of gas

{X}_{2} \to

Mole fraction of gas in solution

{K}_{H} =

Henry's law constant
So,

p ∝

solubility

\dfrac{{p}_{1}}{{p}_{2}} = \dfrac{{s}_{1}}{{s}_{2}} ⇒ \dfrac{500}{0 . 01} = \dfrac{750}{x}

∴ x = 0 . 015g / L

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