The solubility of silver chromate is 1.992 ti | vedclass

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(B) $1$. Calculate the molar solubility $(S)$ of $Ag_2CrO_4$ in $mol/L$: $S = \frac{	ext{solubility in } g/L}{	ext{molar mass in } g/mol} = \frac{1.

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$1.2 	imes 10^{-4} \ M$

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(B) $1$. Calculate the molar solubility $(S)$ of $Ag_2CrO_4$ in $mol/L$: $S = \frac{	ext{solubility in } g/L}{	ext{molar mass in } g/mol} = \frac{1.992 	imes 10^{-2}}{332} = 6 	imes 10^{-5} \ mol/L$. $2$. The dissociation of silver chromate is: $Ag_2CrO_4(s) ightleftharpoons 2Ag^{+}(aq) + CrO_4^{2-}(aq)$. $3$. From the stoichiometry,the concentration of $Ag^{+}$ is $2 	imes S$. $4$. $[Ag^{+}] = 2 	imes (6 	imes 10^{-5} \ M) = 1.2 	imes 10^{-4} \ M$.

The solubility of silver chromate is $1.992 \times 10^{-2} \ g/L$. The molar mass of $Ag_2CrO_4$ is $332 \ g/mol$. What is the molar concentration of $Ag^{+}$ in a saturated solution of $Ag_2CrO_4$?

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