Verified Solution • 2025–26 Session moderate

QJEEM22Chbf80: JEE Main 2022 Chemistry - Class 11 JEEM22Chbf80 Solution: The solubility product of a sparingly soluble salt A 2 X 3 is…

The solubility product of a sparingly soluble salt A2X3 is 1.1 ×\times 10 −- 23. If specific conductance of the solution is 3 ×\times 10 −- 5 S m −- 1, the limiting molar conductivity of the solution is x ×x \,\times 10 −- 3 S m2 mol −- 1. The value of x is ___________.

Step-by-Step Verified Solution

A2X3⇌2 A2S+3X3SA_{2} X_{3} \rightleftharpoons \underset{2S}{2 \mathrm{~A}}+\underset{3S}{3 \mathrm{X}}

amp;Ksp=(2 s)2(3s)3=1.1×10−23amp; S≈10−5\begin{aligned}&\mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^{2}(3 s)^{3}=1.1 \times 10^{-23} \\\\&\mathrm{~S} \approx 10^{-5}\end{aligned}

For sparingly soluble salts

∧mamp;=∧m0∧mamp;=kS×103amp;=3×10−510−5×10−3amp;=3×10−3 Sm2 mol−1\begin{aligned}\wedge_{m} &=\wedge_{m}^{0} \\\\\wedge_{m} &=\frac{\mathrm{k}}{\mathrm{S} \times 10^{3}} \\\\&=\frac{3 \times 10^{-5}}{10^{-5}} \times 10^{-3} \\\\&=3 \times 10^{-3} ~ \mathrm{Sm}^{2} \mathrm{~mol}^{-1}\end{aligned}

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