Question

What is the maximum volume of water required to dissolve 2g of calcium sulphate at 298K ? $(\ce{K_{sp}}$ for $\ce{CaSO_4}$ is $9.0 \times 10^{-6})$

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

Answer 1

Answer: (C)

Given,
Solubility product for $CaSO_4\, (K_{sp}) = 9.0 \times 10^{-6}$
Gram value of $CaSO_4 =2\,g$
Suppose solubility of $CaSO_4$ in water is ‘$S$’.
$\ce{CaSO_4 <=>}$$ \underset{\text{(S)}}{\ce{\overset{2+}{Ca}}}\ce{(aq) +}$ $\underset{\text{(S)}}{\ce{SO^{2-}4}}\ce{(aq)}$
Here, $K_{sp} (CaSO_4)=S^2$
$S^2=9.0 \times 10^{-6}$
$S=3 \times 10^{-3}$
Molar mass of $CaSO_4 =136$
Solubility of $CaSO_4 =3 \times 10^{-3}$
When, $2g$ of $CaSO_4$ dissolve in water.
$\because$ Solubility $=\frac{2}{136 \times \text{Volume}}$
$\therefore 3 \times1 0^{-3}=\frac{2}{136 \times \text{Volume}}$
Volume $=\frac{2}{136 \times 3 \times10^{-3}}$
$=\frac{2\times1000}{136\times 3}=4.90\,L$