(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
M.ww×VL1=2×10−3
2780.1×VL1=2×10−3
VL=278×20.1×1000=0.18L
The minimum volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution ( K_sp of PbCl_2=3.2× 10^-8 ;atomic mass of…
JEE Main 2018 — Chemistry Physical Chemistry
2018mcqmedium
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=207u ) is:
Official previous-year question
Held on 15 Apr 2018 · Verified 6 Jul 2026.
Options
- A
1.798 L
- B
0.36 L
- C
17.98 L
- D
0.18 L
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