$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9 \times 10^{-6}$. The minimum volume of water needed…
$\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9 \times 10^{-6}$. The minimum volume of water needed to dissolve $1 \mathrm{~g}$ of $\mathrm{CaSO}_4$ at $298 \mathrm{~K}$ temperature is ......
$3.50 \mathrm{~L}$
$4.25 \mathrm{~L}$
$1.75 \mathrm{~L}$
$2.45 \mathrm{~L}$
Solution
Let $S$ be the solubility of $\mathrm{CaSO}_4$.
$
\begin{aligned}
& \mathrm{CaSO}_4 \rightleftharpoons \mathrm{Ca}^{2+}+\mathrm{SO}_4^{2-} \\
& {\left[\mathrm{Ca}^{2+}\right]=\left[\mathrm{SO}_4^{2-}\right]=S } \\
& K_{\mathrm{sp}}\left[\mathrm{Ca}^{2+}\right]\left[\mathrm{SO}_4^{2-}\right]=S \times S=S^2=9 \times 10^{-6} \\
\Rightarrow \quad & S=0.003 \mathrm{M}
\end{aligned}
$
The molar mass of $\mathrm{CaSO}_4$ is
$
40+32+64=136 \mathrm{~g}
$
The solubility in $\mathrm{g} / \mathrm{L}$ is $0.003 \times 136=0.408 \mathrm{~g} / \mathrm{L}$ means, $0.408 \mathrm{~g}$ dissolves in $1 \mathrm{~L}$.
$1 \mathrm{~g}$ will dissolve in $=1 / 0.408=2.45 \mathrm{~L}$
Hence, the correct option is (4)