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The minimum volume of water required to dissolve 1 g of calcium sulphate at 298 K is ( $K_{s p}$ is $9.1 \times 10^{- 6}$ )

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2.45 L

4 L

10 L

5 L

answer is A.

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Detailed Solution

$C a S O_{4}$

$K_{s p} = S^{2} = 9.1 \times 10^{- 6}$

$S = 3 \times 10^{- 3} m o l e / L$

Solubility in grams $= 3 \times 10^{- 3} \times 136 g / L$

$= 0.408 g / L$

$0.408 g \to 1 L$

$1 g \to \frac{1}{0.408} = 2.45 L$

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