Answer:
Let solubility of \[PbC{{l}_{2}}\] is
's' \[mol\,{{L}^{-1}}\]
\[\underset{s}{\mathop{PbC{{l}_{2}}}}\,\to
\underset{s}{\mathop{P{{b}^{2+}}}}\,+\underset{2s}{\mathop{2C{{l}^{-}}}}\,\]
\[{{K}_{sp}}=[P{{b}^{2+}}]{{[C{{l}^{-}}]}^{2}}\]
\[=s\times
{{(2s)}^{2}}=4{{s}^{3}}\]
\[s={{\left[
\frac{{{K}_{sp}}}{4} \right]}^{\frac{1}{3}}}={{\left[ \frac{3.2\times
{{10}^{-8}}}{4} \right]}^{\frac{1}{3}}}\]
\[={{[8\times
{{10}^{-9}}]}^{\frac{1}{3}}}=2\times {{10}^{-3}}mol\,{{L}^{-1}}\]
Solubility in g \[{{L}^{-1}}\]
molar mass of \[PbC{{l}_{2}}\]
\[=s\times 278\]
\[=2\times {{10}^{-3}}\times 278\]
\[=0.556g\,{{L}^{-1}}\]
Volume of water required to
dissolve 0.1 g \[PbC{{l}_{2}}\]
\[=\frac{1000\times 0.1}{0.556}mL\]
\[=179.8mL\]
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