The solubility product of AgCl is 10^{-10} , | vedclass

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(D) The solubility $S$ of $AgCl$ is given by $K_{sp} = S^2$. $S = \sqrt{10^{-10}} = 10^{-5} \, mol/L$. Molar mass of $AgCl = 143.5 \, g/mol$. Mass of

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$0.01$

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(D) The solubility $S$ of $AgCl$ is given by $K_{sp} = S^2$. $S = \sqrt{10^{-10}} = 10^{-5} \, mol/L$. Molar mass of $AgCl = 143.5 \, g/mol$. Mass of $AgCl = 14.35 \, mg = 14.35 	imes 10^{-3} \, g$. Number of moles of $AgCl = \frac{14.35 	imes 10^{-3} \, g}{143.5 \, g/mol} = 10^{-4} \, mol$. Since $S = \frac{n}{V}$,we have $V = \frac{n}{S} = \frac{10^{-4} \, mol}{10^{-5} \, mol/L} = 10 \, L$. Converting $10 \, L$ to $m^3$: $10 \, L = 10 	imes 10^{-3} \, m^3 = 0.01 \, m^3$.

The solubility product of $AgCl$ is $10^{-10} \, M^2$. The minimum volume (in $m^3$) of water required to dissolve $14.35 \, mg$ of $AgCl$ is approximately

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