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The solublity of $\mathrm{BaSO}_{4}$ in water $2.42 \times 10^{3}\; \mathrm{gL}^{-1}$ at $298 \;\mathrm{K} .$ The value of solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ will be (Given molar mass of $\mathrm{BaSO}_{4}=233\; \mathrm{g} \;\mathrm{mol}^{-1}$ )
$1.08 \times 10^{-10}\; \mathrm{mol}^{2} \;\mathrm{L}^{-2}$
$1.08 \times 10^{-12}\; \mathrm{mol}^{2} \;\mathrm{L}^{-2}$
$1.08 \times 10^{-14}\; \mathrm{mol}^{2} \;\mathrm{L}^{-2}$
$1.08 \times 10^{-8}\; \mathrm{mol}^{2} \;\mathrm{L}^{-2}$
Solution
solublity of $\mathrm{BaSO}_{4}=2.42 \times 10^{-3} \mathrm{gL}^{-1}$
$\therefore \mathrm{s}=\frac{2.42 \times 10^{-3}}{233}=1.038 \times 10^{-5} \mathrm{mol} \mathrm{L}^{-1}$
$\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^{2} =\left(1.038 \times 10^{-5}\right)^{2}$
$=1.08 \times 10^{-10} \mathrm{mol}^{2} \mathrm{L}^{-2}$
