Q29 / 90 · JEE Main 2021 (31 Aug Shift 1) · Chemistry

$A_{3} B_{2}$ is a sparingly soluble salt of molar mass $M (g {mol}^{- 1})$ and solubility $x g L^{- 1} .$ The solubility product satisfies $K_{sp} = a {(\frac{x}{M})}^{5} .$ The value of $a$ is __________ . (Integer answer)

Correct answer: 108

Solution

Solubility

= (\frac{x}{m})

mole/lit

A_{3} B_{2} (s) \underset{\frac{3 x}{M}}{⇌ 3 A^{2 +}} \underset{2 (\frac{x}{M})}{(aqs) + 2 B^{3 -} (aqs)}

K_{sp} = {(\frac{3 x}{M})}^{3} {(\frac{2 x}{M})}^{2}

= (3)^{3} {(2)}^{2} {(\frac{x}{m})}^{5}

= 27 \times 4 {(\frac{x}{m})}^{5}

= 108 {(\frac{x}{m})}^{5} = a {(\frac{x}{M})}^{5}

So

a = 108

Hence, the correct answer is 108.

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