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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium

JEE Main 2021 · 31 August · Morning Shift · Chemistry · Ionic Equilibrium

A _ 3 B _ 2 is a sparingly soluble salt of molar mass M(g mol ^ - 1...

{A}_{3}{B}_{2}isasparinglysolublesaltofmolarmassis a sparingly soluble salt of molar massM(gmol−1)M\left(g{mol}^{- 1}\right) and solubility xgL−1.xg{L}^{- 1} . The solubility product satisfies Ksp=a(xM)5.{K}_{sp} = a{\left(\dfrac{x}{M}\right)}^{5} . The value of aa is __________ . (Integer answer)

Correct answer

0

Step-by-step solution

Solubility =\left(\dfrac{x}{m}\right)mole/lit<br/>mole/lit<br />{A}_{3}{B}_{2}\left(s\right)\mathop{⇌ 3{A}^{2 +}}_{\dfrac{3x}{M}}\mathop{( aqs ) + 2{B}^{3 -} ( aqs )}_{2\left(\dfrac{x}{M}\right)}<br/><br />{K}_{sp} = {\left(\dfrac{3x}{M}\right)}^{3}{\left(\dfrac{2x}{M}\right)}^{2}<br/><br />= \left(3\right){}^{3}{\left(2\right)}^{2}{\left(\dfrac{x}{m}\right)}^{5}<br/><br />= 27 \times 4{\left(\dfrac{x}{m}\right)}^{5}<br/><br />= 108{\left(\dfrac{x}{m}\right)}^{5} = a{\left(\dfrac{x}{M}\right)}^{5}<br/>So<br />Soa = 108
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