JEE Main 2026 · 5 Apr · Shift 1 · Chemistry Q55

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ChemistrySingle correct+4 / -1Question 55 of 75

$\ce{M3A2}$ is a sparingly soluble salt of molar mass $y$ g mol $^{-1}$ and solubility $x$ g L $^{-1}$ . The ratio of the molar concentration of the anion ( $\ce{A^{3-}}$ ) to the solubility product of the salt is

\dfrac1{54}\cdot\dfrac{y^4}{x^4}

\dfrac{y^5}{108x^4}

108\cdot\dfrac{x^5}{y^5}

\dfrac1{108}\dfrac{y^4}{x^4}

Answer and worked solution

Official answer: A

Solubility. $s=\dfrac xy$ mol L $^{-1}$ . $\ce{M3A2 <=> 3M^{2+} + 2A^{3-}}$ , so $[\ce{A^{3-}}]=2s$ .

Solubility product. $K_{sp}=(3s)^3(2s)^2=108s^5$ .

Ratio.
$\frac{2s}{108s^5}=\frac1{54s^4}=\frac1{54}\cdot\frac{y^4}{x^4}.$

(A)

Topics:Solubility product

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JEE Main 2026 · 5 Apr · Shift 1, Chemistry question 55