Question

What is the minimum volume of water required to dissolve $1g$ of calcium sulphate at $298K?K_{sp}$ for $CaS{O}_4$ is $9.0\times 10^{-6}$ .
(Molar mass of $CaS{O}_4=136gmo{l}^{-1})$

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

Answer 1

Answer: (A)

$CaS{O}_4\to C{a}^{2+}+S{O}_4^{2-}$
$K_{sp}=[C{a}^{2+}][S{O}_4^{2-}]$
$K_{sp}=S^2$
$9.0\times 10^{-6}=S^2$
$S=\sqrt{9.0\times 10^{-6}}$
$=3.0\times 10^{-3}mol/L$
Given, molecular mass $=136gmo{l}^{-1}$
Solubility of $CaS{O}_4=3.0\times 10^{-3}\times 136g/L$
$=0.14g/L$
$\therefore$ To dissolve $0.41$ of $CaS{O}_4$, water requires $=1L$
$\therefore$ To dissolve $1g$ of $CaS{O}_4$, water requires $=\frac{1}{0.41}L$
$=2.439L=2.44L$