Verified Solution • 2025–26 Session moderate

QJEEM18Ch6421: JEE Main 2018 Chemistry - Class 11 JEEM18Ch6421 Solution: The minimum volume of water required to dissolve 0.1 g lead…

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

Multiple Choice Options

  • A 0.36 L
  • B 17.98 L
  • C 0.18 L
  • D 1.798 L

Step-by-Step Verified Solution

Given,

Ksp of PbCl2 = 3.2 ×\times 10 −- 8

PbCl2 ⇌\rightleftharpoons Pb2+ 2Cl
Initially 1 0 0
At equilibrium 1-s s 2s


Ksp = [s] [2s]2

∴    \therefore\,\,\,\, Ksp = 4s3

∴    \therefore\,\,\,\, 4s3 = 3.2 ×\times 10 −- 8

⇒\Rightarrow     \,\,\,\, s3 = 8 ×\times 10 −- 9

⇒\Rightarrow     \,\,\,\, s = 2 ×\times 10 −- 3 M

Solubility = no.  of moles  of  pbcl2 volume (in  litre){{no.\,\,of\,moles\,\,of\,\,pbc{l_2}\,} \over {volume\,(in\,\,litre)}}

∴    \therefore\,\,\,\, 2 ×\times 10 −- 3 = 0.1278{{0.1} \over {278}} ×\times 1V{1 \over V}

⇒\Rightarrow     \,\,\,\, V = 0.1278{{0.1} \over {278}} ×\times 1032{{{{10}^3}} \over 2} = 0.18 L

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring