K_{sp} for CaSO_4 is 9 times 10^{-6}. The min | vedclass

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(D) Let $S$ be the solubility of $CaSO_4$ in $mol/L$.$CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$$K_{sp} = [Ca^{2+}][SO_4^{2-}] = S^2 =

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$2.45$

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(D) Let $S$ be the solubility of $CaSO_4$ in $mol/L$.$CaSO_4(s) ightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$$K_{sp} = [Ca^{2+}][SO_4^{2-}] = S^2 = 9 	imes 10^{-6}$$S = \sqrt{9 	imes 10^{-6}} = 3 	imes 10^{-3} \ M$ (or $mol/L$).The molar mass of $CaSO_4 = 40 + 32 + (4 	imes 16) = 136 \ g/mol$.Solubility in $g/L = S 	imes 	ext{Molar Mass} = 3 	imes 10^{-3} \ mol/L 	imes 136 \ g/mol = 0.408 \ g/L$.This means $0.408 \ g$ of $CaSO_4$ dissolves in $1 \ L$ of water.Therefore,the volume required to dissolve $1 \ g$ of $CaSO_4 = \frac{1 \ g}{0.408 \ g/L} \approx 2.45 \ L$.Hence,the correct option is $D$.

$K_{sp}$ for $CaSO_4$ is $9 \times 10^{-6}$. The minimum volume of water needed to dissolve $1 \ g$ of $CaSO_4$ at $298 \ K$ temperature is ...... (in $L$)

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