JEE Main 2022 — Chemical 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
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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