Q. The minimum volume of water required to dissolve $0.1g$ lead (II) chloride to get a saturated solution is:
$\left(\right.K_{sp}of\left(PbCl\right)_{2}=3.2\times \left(10\right)^{- 8};\text{atomic mass of}Pb=207u\left.\right)$

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$0.36L$

$0.18L$

$17.98L$

$1.798L$

Solution:

$PbCl_{2}\rightleftharpoonsPb^{+ 2}+2Cl^{-}s2s$
$4s^{3}=3.2\times 10^{- 8}$
$s^{3}=8\times 10^{- 9}$
$s=2\times 10^{- 3}$
$s=\frac{0 . 1}{278}\times \frac{1000}{ V}\Rightarrow V=180ml$

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