JEE Main 2021ChemistrySolutionsMediumNumerical

JEE Main 2021 — Solutions Question with Solution

JEE Main 2021 (17 Mar Shift 1)

Question

The oxygen dissolved in water exerts a partial pressure of 20 kPa in the vapour above water. The molar solubility of oxygen in water is ______ ×10-5 mol dm-3.

(Round off to the Nearest Integer).

[Given : Henry's law constant =KH=8.0×104 kPa for O2. Density of water with dissolved oxygen=1.0 kg dm-3]

Enter your answer

Show full solutionCorrect answer: 1389
Correct answer
1389

Step-by-step explanation

P=KH·x

or, 20×103=8×104×103×nO2nO2+nwater 

or, 14000=nO2nO2+nwater =nO2nwater

Means 1 mole water =18 gm=18 ml dissolves

14000 moles O2. Hence, molar solubility

=1400018×1000=172 mol dm-3

=1388.89×10-5 mol dm-3≈1389 ×10-5 mol dm-3

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About this question

This is a previous-year question from JEE Main 2021, covering the Solutions chapter of Chemistry. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.