The solubility product of AgCl is 1.5625 time | vedclass

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Question

(D) The solubility $S$ in $mol/L$ is given by $S = \sqrt{K_{sp}}$.Given $K_{sp} = 1.5625 	imes 10^{-10}$.$S = \sqrt{1.5625 	imes 10^{-10}} = 1.25

Vedclass · Accepted Answer

$1.79 	imes 10^{-3}$

Vedclass · Answer

(D) The solubility $S$ in $mol/L$ is given by $S = \sqrt{K_{sp}}$.Given $K_{sp} = 1.5625 	imes 10^{-10}$.$S = \sqrt{1.5625 	imes 10^{-10}} = 1.25 	imes 10^{-5} \ mol/L$.The molar mass of $AgCl$ is $107.87 + 35.45 = 143.32 \ g/mol$ (approximately $143.5 \ g/mol$).Solubility in $g/L = S 	imes 	ext{Molar Mass} = 1.25 	imes 10^{-5} \ mol/L 	imes 143.5 \ g/mol = 1.79375 	imes 10^{-3} \ g/L$.Thus,the solubility is approximately $1.79 	imes 10^{-3} \ g/L$.

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

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