At 298 K temperature,the solubility product | vedclass

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(B) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility be $S \ mol \ L^{-1}$. Then $K_{sp} =

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$1.75 	imes 10^{-3} \ g \ L^{-1}$

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(B) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.Let the solubility be $S \ mol \ L^{-1}$. Then $K_{sp} = [Ag^+][Cl^-] = S^2$.$S = \sqrt{K_{sp}} = \sqrt{1.5 	imes 10^{-10}} = 1.2247 	imes 10^{-5} \ mol \ L^{-1}$.To convert solubility from $mol \ L^{-1}$ to $g \ L^{-1}$,multiply by the molar mass $(143.5 \ g \ mol^{-1})$:Solubility in $g \ L^{-1} = 1.2247 	imes 10^{-5} \ mol \ L^{-1} 	imes 143.5 \ g \ mol^{-1} = 1.757 	imes 10^{-3} \ g \ L^{-1}$.

At $298 \ K$ temperature,the solubility product $(K_{sp})$ for $AgCl$ is $1.5 \times 10^{-10}$. Calculate its solubility in $g \ L^{-1}$ in pure water. (Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

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