Q6 / 75 · JEE Main 2026 (02 April Shift 1) · Chemistry
The solubility product constants of $Ag_2CrO_4$ and $AgBr$ are $32x$ and $4y$ respectively at $298$ K. The value of $\left(\dfrac{\text{molarity of } Ag_2CrO_4}{\text{molarity of } AgBr}\right)$ can be expressed as :
- A$\dfrac{2\sqrt[3]{x}}{y}$
- B$2\sqrt{\dfrac{x}{y}}$
- C$\sqrt{\dfrac{x}{y}}$
- Correct answerD$\dfrac{\sqrt[3]{x}}{\sqrt{y}}$
Solution
Let the solubility of $Ag_2CrO_4$ be $S_1$.
For $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$
$K_{sp} = (2S_1)^2(S_1) = 4S_1^3$
$4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = 2\sqrt[3]{x}$
Let the solubility of $AgBr$ be $S_2$.
For $AgBr \rightleftharpoons Ag^+ + Br^-$
$K_{sp} = S_2^2$
$S_2^2 = 4y \Rightarrow S_2 = 2\sqrt{y}$
The ratio of their molarities is:
$\dfrac{S_1}{S_2} = \dfrac{2\sqrt[3]{x}}{2\sqrt{y}} = \dfrac{\sqrt[3]{x}}{\sqrt{y}}$
Answer: $\dfrac{\sqrt[3]{x}}{\sqrt{y}}$
For $Ag_2CrO_4 \rightleftharpoons 2Ag^+ + CrO_4^{2-}$
$K_{sp} = (2S_1)^2(S_1) = 4S_1^3$
$4S_1^3 = 32x \Rightarrow S_1^3 = 8x \Rightarrow S_1 = 2\sqrt[3]{x}$
Let the solubility of $AgBr$ be $S_2$.
For $AgBr \rightleftharpoons Ag^+ + Br^-$
$K_{sp} = S_2^2$
$S_2^2 = 4y \Rightarrow S_2 = 2\sqrt{y}$
The ratio of their molarities is:
$\dfrac{S_1}{S_2} = \dfrac{2\sqrt[3]{x}}{2\sqrt{y}} = \dfrac{\sqrt[3]{x}}{\sqrt{y}}$
Answer: $\dfrac{\sqrt[3]{x}}{\sqrt{y}}$
Hence, the correct answer is option: 'D'.