Answer:
Let
solubility of \[CaS{{O}_{4}}\] at
298 K is s\[mol\,{{L}^{-1}}\].
Then \[{{K}_{sp}}={{s}^{2}}\]
\[s=\sqrt{9.1\times {{10}^{-6}}}=3.016\times
{{10}^{-3}}mol/litre\]
Solubility in \[g/L\,=3.016\times {{10}^{-3}}\times
136=0.41\,g/L\]
\[\therefore \] Minimum volume of water required to
dissolve
\[1\,g\,CaS{{O}_{4}}=\frac{1000}{0.41}=2439\,mL\]
\[=2.439\,L\]
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