JEE Main 2018ChemistryIonic EquilibriumMediumMCQ

JEE Main 2018 — Ionic Equilibrium Question with Solution

JEE Main 2018 (15 Apr)

Question

The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( Ksp of PbCl2=3.2×10-8 ;atomic mass of Pb=207 u ) is:

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Show full solutionCorrect option: D
Correct answer
D0.18 L

Step-by-step explanation

(Ksp)PbCl2=32×10-9

PbCl2=Pb2+  s   +2Cl- 2s

Ksp= Pb2+ 2Cl-2

Ksp=4s3 = 32×10-9

s3 = 8×10-9

s=2×10-3M

wM.w×1VL=2×10-3

0.1278×1VL=2×10-3

VL=0.1×1000278×2=0.18 L 

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About this question

This is a previous-year question from JEE Main 2018, covering the Ionic 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.