JEE Main 2019ChemistrySolutionsMediumMCQ

JEE Main 2019 — Solutions Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg , lowering of vapour pressure will be:
(molar mass of urea =60 g mol-1 )

Choose an option

Show full solutionCorrect option: D
Correct answer
D0.017 mm Hg

Step-by-step explanation

Relative lowering in vapour Pressure
⇒ Mole of urea nB=0.660=10-2 mole ;

Mole of water nA=36018=20

Po-PsPo=xB=Po-PsPo=nBnn+nB

Here =Po=V.P. of pure solvent

Ps=V.P of solution.

nA+nB≃nA

Po-PsPo=nBnA
lowering of vapour pressure(Po-Ps)=nBnA×Po

Po-Ps=10-220×35

Po-Ps=0.0175  mm of Hg

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About this question

This is a previous-year question from JEE Main 2019, 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.