Question

What is the minimum volume of water required to dissolve $ 1\, g $ of calcium sulphate at $ 298 K? K_{sp} $ for $ CaSO_{4} $ is $ 9.0 \times 10^{-6} $ .
(Molar mass of $ CaSO_4 = 136 \,g \,mol ^{-1}) $

• A. $ 2.45\,L $
• B. $ 4.08\,L $
• C. $ 4.90\,L $
• D. $ 3.00\,L $

Answer 1

Answer: (A)

$CaSO_4 \rightarrow Ca^{2+} + SO_4^{2-}$
$K_{sp} = [Ca^{2+}][SO_4^{2-}]$
$ K_{sp} = S^2$
$ 9.0 \times 10^{-6} = S^2$
$S = \sqrt{9.0 \times 10^{-6}}$
$ = 3.0 \times 10^{-3}\,mol/L$
Given, molecular mass $= 136\,g \,mol^{-1}$
Solubility of $CaSO_4 = 3.0 \times 10^{-3} \times 136 \,g/L$
$ = 0.14\, g/L$
$\therefore $ To dissolve $0.41$ of $CaSO_4$, water requires $= 1\, L$
$\therefore $ To dissolve $1g$ of $CaSO_4$, water requires $ = \frac{1}{0.41}L$
$ = 2.439\,L = 2.44\,L$