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Chemistry · Ionic Equilibrium

JEE Main 2026 — 2 April, Morning Shift — Question 45

The solubility product constants of Ag2CrO4\mathrm{Ag}_{2} \mathrm{CrO}_{4} and AgBr are 32x and 4y respectively at 298 K .

The value of ( molarity of Ag2CrO4 molarity of AgBr)\left(\frac{\text { molarity of } \mathrm{Ag}_{2} \mathrm{CrO}_{4}}{\text { molarity of } \mathrm{AgBr}}\right) can be expressed as :

  1. Option A:

    2x3y\frac{2 \sqrt[3]{x}}{y}

  2. Option B:

    2xy2 \sqrt{\frac{x}{y}}

  3. Option C:

    xy\sqrt{\frac{x}{y}}

  4. Option D:

    x3y\frac{\sqrt[3]{x}}{\sqrt{y}}

    Correct

Answer: D

Step-by-step solution

For Ag2C2O4\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4} 4 S13=32x4 \mathrm{~S}_{1}{ }^{3}=32 \mathrm{x} ⇒S1=2x1/3\Rightarrow \mathrm{S}_{1}=2 \mathrm{x}^{1 / 3} For AgBr S22=4yS_{2}{ }^{2}=4 y ⇒S2=2y1/2\Rightarrow \mathrm{S}_{2}=2 \mathrm{y}^{1 / 2} Ratio (S1S2)=x3y\left(\frac{S_{1}}{S_{2}}\right)=\frac{\sqrt[3]{x}}{\sqrt{y}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions

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