35. Calculate the volume of water required to dissolve 0.1 g lead (II) chloride to get a saturated solution.

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Suppose, solubility of PbCl2 in water is s mol L-1
$\underset{\left(1-\mathrm{s}\right)}{{\mathrm{PbCl}}_{2}\left(\mathrm{s}\right)}⇌\underset{\mathrm{s}}{{\mathrm{Pb}}^{2+}\left(\mathrm{aq}\right)}+\underset{2\mathrm{s}}{2{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{sp}}=\left[{\mathrm{Pb}}^{2+}\right].{\left[{\mathrm{Cl}}^{-}\right]}^{2}\phantom{\rule{0ex}{0ex}}{\mathrm{K}}_{\mathrm{sp}}=\left[\mathrm{s}\right]{\left[2\mathrm{s}\right]}^{2}=4{\mathrm{s}}^{3}\phantom{\rule{0ex}{0ex}}32×{10}^{-8}=4{\mathrm{s}}^{3}\phantom{\rule{0ex}{0ex}}{\mathrm{s}}^{3}=\frac{3.2×{10}^{-8}}{4}=0.8×{10}^{-8}\phantom{\rule{0ex}{0ex}}{\mathrm{s}}^{3}=8.0×{10}^{-9}$
Solubility of
Solubility of
($\because$Molar mass of ${\mathrm{PbCl}}_{2}=207+\left(2×35.5\right)=278$)
0.556 g of PbCl2 dissolve in 1 L of water.
will dissolve in
To make a saturated solution, dissolution of 0.1 g PbCl2 in  of water will be required.