M 3 A 2 is a sparingly soluble salt of molar mass y g mol -1 and solubility × g L -1 . The ratio of the molar concentration of the anion ( A 3- ) to the solubility product of the salt is

Chemistry · Solutions and Colligative Properties

JEE Main 2026 — 5 April, Morning Shift — Question 53

$\mathrm{M}_{3} \mathrm{~A}_{2}$ is a sparingly soluble salt of molar mass y $\mathrm{g} \mathrm{mol}^{-1}$ and solubility $\times \mathrm{g} \mathrm{L}^{-1}$ . The ratio of the molar concentration of the anion ( $\mathrm{A}^{3-}$ ) to the solubility product of the salt is

Option A:
$\frac{1}{54} \cdot \frac{y^{4}}{x^{4}}$
Correct
Option B:
$\frac{y^{5}}{108 x^{4}}$
Option C:
$108 \cdot \frac{x^{5}}{y^{5}}$
Option D:
$\frac{1}{108} \frac{y^{4}}{x^{4}}$

Answer: A

Step-by-step solution

Molarity

\begin{array}{ccccc} \mathrm{M_3A_2}(s) & \rightleftharpoons & 3\mathrm{M^{2+}} & + & 2\mathrm{A^{3-}} \\ & & \dfrac{3x}{y} & & \dfrac{2x}{y} \end{array}

$\mathrm{K_{sp}} = [\mathrm{M^{2+}}]^3[\mathrm{A^{3-}}]^2$

$\mathrm{K_{sp}} = \left[\frac{3x}{y}\right]^3\left[\frac{2x}{y}\right]^2$

Ratio of $\dfrac{[\mathrm{A^{3-}}]}{\mathrm{K_{sp}}}$

$= \frac{\dfrac{2x}{y}}{\left[\dfrac{3x}{y}\right]^3\left[\dfrac{2x}{y}\right]^2}$

$= \frac{1}{54}\frac{y^4}{x^4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Solid in Liquid Solutions (Colligative Properties)