JEE Main 2022ChemistryChemical EquilibriumChemical Equilibrium Law Of Mass Action And Equilibrium ConstantmediumNumerical

JEE Main 2022Chemical Equilibrium Question with Solution

From: JEE Main 2022 (Online) 29th June Evening Shift

Question

A box contains 0.90 g of liquid water in equilibrium with water vapour at 27C. The equilibrium vapour pressure of water at 27C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]

(Given : R = 0.082 L atm K1 mol1)

(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

Enter your answer

Show full solutionCorrect answer: 29
Correct answer
29

Step-by-step explanation

We know, 760 Torr = 1 atm

32 Torr = atm

As all the liquid water evaporates so entire water is in gaseous state.

Weight of water vapour = 0.9 g

Moles of water vapour (n) =

Pressure (P) = atm

Temperature (T) = (27 + 273) K = 300 K

R = 0.082 L atm K1 mol1

Given water vapour act as an ideal gas, so we can apply ideal gas equation.

From ideal gas equation,

PV = nRT

L

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About this question

This is a previous-year question from JEE Main 2022, covering the Chemical Equilibrium 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.