JEE Main 2026 · 2 Apr · Shift 1 · Chemistry Q56

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ChemistrySingle correct+4 / -1Question 56 of 75

The solubility product constants of $\ce{Ag2CrO4}$ and AgBr are $32x$ and $4y$ respectively at 298 K. The value of $\left(\dfrac{\text{molarity of }\ce{Ag2CrO4}}{\text{molarity of AgBr}}\right)$ can be expressed as :

\dfrac{2\sqrt[3]x}y

2\sqrt{\dfrac xy}

\sqrt{\dfrac xy}

\dfrac{\sqrt[3]x}{\sqrt y}

Answer and worked solution

Official answer: D

Solubilities.

$\ce{Ag2CrO4}$ : $4s_1^3=32x$ , so $s_1=2x^{1/3}$ .
AgBr: $s_2^2=4y$ , so $s_2=2\sqrt y$ .

Ratio. $\dfrac{s_1}{s_2}=\dfrac{\sqrt[3]x}{\sqrt y}$ .

(D)

Topics:Solubility product

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JEE Main 2026 · 2 Apr · Shift 1, Chemistry question 56