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)

$2.45\,L$

Answer 2

$CaSO_4 {->} Ca^{2+} + SO_4^{2-}$
$K_{sp} = [Ca^{2+}][SO_4^{2-}]$
$K_{sp} = S^2$
$9.0 \times 10^{-6} = S^2$
$S = \sqrt{9.0 \times 10^{-6}}$
$= 3.0 \times 10^{-3}\,mol/L$
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 $= \frac{1}{0.41}L$
$= 2.439\,L = 2.44\,L$