The solubility of silver sulphate in the water at $100^{o}$ C is approximately $1.4$ g per $100$ mL. What is the solubility product of this salt at $100^{o}$ C?

5.7 \times 10^{- 8}

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3.5 \times 10^{- 7}

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8.3 \times 10^{- 6}

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4.1 \times 10^{- 5}

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3.6 \times 10^{- 4}

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Solution

The correct option is E $3.6 \times 10^{- 4}$
The molecular weight of $A g_{2} S O_{4}$ is $312$ g/mol.
Solubility $S = 1.4 g / 100 m l = 14 g / L = \frac{14}{312} M = 0.04487 M$
The expression for the solubility product is $K_{S P} = [A g^{+}]^{2} [S O_{4}^{2 -}] = (2 S)^{2} (S) = 4 S^{3}$
Substitute $S = 0.04487 M$ in the above expression.
$K_{S P} = 4 S^{3} = 4 (0.04487)^{3} = 3.6 \times 10^{- 4}$

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1.5625 \times 10^{- 10}

25^{o} C

g L^{- 1}

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The solubility product of $M g (O H)_{2}$ at $25^{\circ} C$ is $1.4 \times 10^{- 11}$ . The solubility of $M g (O H)_{2}$ in gm per litre is:

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