The solubility of AgCl at 20 , ^oC is 1.435 t | vedclass

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(B) The molar mass of $AgCl$ is $108 + 35.5 = 143.5 \, g/mol$. The solubility in $g/L$ is given as $S = 1.435 	imes 10^{-3} \, g/L$. The molar solubi

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$1 	imes 10^{-10}$

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(B) The molar mass of $AgCl$ is $108 + 35.5 = 143.5 \, g/mol$. The solubility in $g/L$ is given as $S = 1.435 	imes 10^{-3} \, g/L$. The molar solubility $(S)$ in $mol/L$ is calculated as: $S = \frac{1.435 	imes 10^{-3} \, g/L}{143.5 \, g/mol} = 10^{-5} \, mol/L$. For $AgCl$,the dissociation is $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$. The solubility product is $K_{sp} = [Ag^+][Cl^-] = S 	imes S = S^2$. Therefore,$K_{sp} = (10^{-5})^2 = 10^{-10}$.

The solubility of $AgCl$ at $20 \, ^oC$ is $1.435 \times 10^{-3} \, g/L$. The solubility product $(K_{sp})$ of $AgCl$ is:

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