(D) For $Ag_2CrO_4$ (sparingly soluble salt of type $A_2B$): $K_{sp} = (2s_1)^2(s_1) = 4s_1^3 = 32x$. Thus,$s_1^3 = 8x$,which gives $s_1 = 2\sqrt[3]{x

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(D) For $Ag_2CrO_4$ (sparingly soluble salt of type $A_2B$): $K_{sp} = (2s_1)^2(s_1) = 4s_1^3 = 32x$. Thus,$s_1^3 = 8x$,which gives $s_1 = 2\sqrt[3]{x}$. For $AgBr$ (sparingly soluble salt of type $AB$): $K_{sp} = s_2^2 = 4y$. Thus,$s_2 = \sqrt{4y} = 2\sqrt{y}$. The ratio of molarities is $\frac{s_1}{s_2} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

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The solubility product constants of $Ag_2CrO_4$ and $AgBr$ are $32x$ and $4y$ respectively at $298 \text{ K}$. The value of $(\frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr})$ can be expressed as :

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A
$\frac{2\sqrt[3]{x}}{y}$
B
$2\sqrt{\frac{x}{y}}$
C
$\sqrt{\frac{x}{y}}$
D
$\frac{\sqrt[3]{x}}{\sqrt{y}}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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The solubility product constants of $Ag_2CrO_4$ and $AgBr$ are $32x$ and $4y$ respectively at $298 \text{ K}$. The value of $(\frac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr})$ can be expressed as :

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