JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

The solubility product constants of $A g_{2} C r O_{4}$ and $A g B r$ are $32 x$ and $4 y$ respectively at $298$ K. The value of $(\frac{molarity of A g _{2} C r O _{4}}{molarity of A g B r})$ can be expressed as :

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

Solution

For $A g_{2} C r O_{4}$ dissolving as $2 A g^{+} + C r O_{4}^{2 -}$ , the solubility product is $K_{s p} = 4 S_{1}^{3}$ . Setting $4 S_{1}^{3} = 32 x$ gives $S_{1} = 2 x^{1/3}$ .

For $A g B r$ (1:1), $K_{s p} = S_{2}^{2} = 4 y$ , so $S_{2} = 2 y^{1/2}$ .

$\frac{S _{1}}{S _{2}} = \frac{2 x ^{1/3}}{2 y ^{1/2}} = \frac{3 x}{y}$

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