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JEE Main 201815 AprilMorning ShiftOnline
Exam
JEE Main
Year
2018
Date
15 April
Shift
Morning Shift
Mode
Online
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2018 · 15 April · Morning Shift · Online · Chemistry · Ionic Equilibrium

The minimum volume of water required to dissolve 0.1 ~g lead (II)...

The minimum volume of water required to dissolve 0.1 g0.1 \mathrm{~g} lead (II) chloride to get a saturated solution \left(K_{\mathrm{SP}}\right. of \mathrm{PbCl}_2=3.2 \times 10^{-8};atomicmassof; atomic mass of\mathrm{Pb}=207 \mathrm{~u}) is :

Options

  1. A1.798 L1.798 \mathrm{~L}
  2. B0.36 L0.36 \mathrm{~L}
  3. C17.95 L17.95 \mathrm{~L}
  4. D0.18 L0.18 \mathrm{~L}

Correct answer

D. 0.18 L0.18 \mathrm{~L}

Step-by-step solution

(Ksp)PbCl2=3.2×10−8=32×10−9PbCl2⇌Pb2+s+2Cl−2sKsp=[Pb2+][Cl−]2 Ksp=4s3=32×10−9 s3=8×10−9 s=2×10−3MwM.W.×1 VL=2×10−30.1278×1 VL=2×10−3 VL=0.1×1000278×2=0.18 L\left(\mathrm{K}_{\mathrm{sp}}\right)_{\mathrm{PbCl}_2}=3.2 \times 10^{-8}=32 \times 10^{-9} \mathrm{PbCl}_2 \rightleftharpoons\underset{\mathrm{s}}{\mathrm{Pb}^{2+}}+\underset{2 \mathrm{s}}{2 \mathrm{Cl}^{-}} \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^2 \mathrm{~K}_{\mathrm{sp}}=4 \mathrm{s}^3=32 \times 10^{-9} \mathrm{~s}^3=8 \times 10^{-9} \mathrm{~s}=2 \times 10^{-3} \mathrm{M} \frac{\mathrm{w}}{\mathrm{M} . \mathrm{W}.} \times \frac{1}{\mathrm{~V}_{\mathrm{L}}}=2 \times 10^{-3} \frac{0.1}{278} \times \frac{1}{\mathrm{~V}_{\mathrm{L}}}=2 \times 10^{-3} \mathrm{~V}_{\mathrm{L}}=\frac{0.1 \times 1000}{278 \times 2}=0.18 \mathrm{~L}
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