Question

What is the maximum volume of water required to dissolve 2g of calcium sulphate at 298K ? $(\mathrm{K}{}_{\mathrm{sp}}$ for $\mathrm{CaSO}{}_4$ 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 $CaS{O}_4(K_{sp})=9.0\times 10^{-6}$
Gram value of $CaS{O}_4=2g$
Suppose solubility of $CaS{O}_4$ in water is ‘$S$’.
$\mathrm{CaSO}{}_4\rightleftharpoons$$\underset{\text{(S)}}{\stackrel{2+}{\mathrm{Ca}}}(\mathrm{aq})+$ $\underset{\text{(S)}}{\mathrm{SO}{}_4^{2-}}(\mathrm{aq})$
Here, $K_{sp}(CaS{O}_4)=S^2$
$S^2=9.0\times 10^{-6}$
$S=3\times 10^{-3}$
Molar mass of $CaS{O}_4=136$
Solubility of $CaS{O}_4=3\times 10^{-3}$
When, $2g$ of $CaS{O}_4$ dissolve in water.
$\because$ Solubility $=\frac{2}{136\times \text{Volume}}$
$\therefore 3\times 10^{-3}=\frac{2}{136\times \text{Volume}}$
Volume $=\frac{2}{136\times 3\times 10^{-3}}$
$=\frac{2\times 1000}{136\times 3}=4.90L$