(B) Step $1$: Calculate molar solubility $(S)$ in $\text{mol/dm}^3$ by dividing solubility in $\text{g/dm}^3$ by molar mass.$S = \frac{18.78 \times 10

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(B) Step $1$: Calculate molar solubility $(S)$ in $\text{mol/dm}^3$ by dividing solubility in $\text{g/dm}^3$ by molar mass.$S = \frac{18.78 \times 10^{-4} \text{ g/dm}^3}{187.8 \text{ g/mol}} = 10^{-5} \text{ mol/dm}^3$.Step $2$: The dissociation of $AB_2$ is $AB_2(s) \rightleftharpoons A^{2+}(aq) + 2B^-(aq)$.Step $3$: The solubility product expression is $K_{sp} = [A^{2+}][B^-]^2 = (S)(2S)^2 = 4S^3$.Step $4$: Substitute $S = 10^{-5}$ into the expression: $K_{sp} = 4(10^{-5})^3 = 4 \times 10^{-15}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

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The solubility of a sparingly soluble salt $AB_2$ is $18.78 \times 10^{-4} \text{ g/dm}^3$. What is its solubility product? (Molar mass of $AB_2 = 187.8 \text{ g mol}^{-1}$)

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A
$2 \times 10^{-15}$
B
$4 \times 10^{-15}$
C
$6 \times 10^{-15}$
D
$8 \times 10^{-15}$

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The solubility of a sparingly soluble salt $AB_2$ is $18.78 \times 10^{-4} \text{ g/dm}^3$. What is its solubility product? (Molar mass of $AB_2 = 187.8 \text{ g mol}^{-1}$)

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