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(B) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility of $AgCl$ in $0.2 \ M \ NaCl$ be $s$.I

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$9 	imes 10^{-10}$

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(B) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility of $AgCl$ in $0.2 \ M \ NaCl$ be $s$.In the presence of $0.2 \ M \ NaCl$,the concentration of $Cl^-$ ions is $[Cl^-] = (s + 0.2) \ M \approx 0.2 \ M$ (since $s$ is very small).The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.Substituting the values: $1.8 	imes 10^{-10} = (s)(0.2)$.Solving for $s$: $s = \frac{1.8 	imes 10^{-10}}{0.2} = 9 	imes 10^{-10} \ M$.

The solubility product of silver chloride at $298 \ K$ is $1.8 \times 10^{-10}$. The solubility of $AgCl$ in $0.2 \ M \ NaCl$ solution will be ............

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