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← JEE Main 2022 · 28 Jun · Shift 1
ChemistryInteger+4 / -1Question 86 of 90

JEE Main 2022 · 28 Jun · Shift 1, Chemistry question 86

The solubility product of a sparingly soluble salt A2X3\mathrm{A_2X_3} is 1.1×10−231.1\times10^{-23}. If specific conductance of the solution is 3×10−53\times10^{-5} S m−1^{-1}, the limiting molar conductivity of the solution is x×10−3x\times10^{-3} S m2^2 mol−1^{-1}. The value of xx is ______.

Answer and worked solution

Official answer: 3

Solubility. Ksp=(2s)2(3s)3=108s5=1.1×10−23K_{sp}=(2s)^2(3s)^3=108s^5=1.1\times10^{-23}, so s≈1.0×10−5s\approx1.0\times10^{-5} mol L−1=10−2^{-1}=10^{-2} mol m−3^{-3}.

Molar conductivity. κc=3×10−510−2=3×10−3\dfrac\kappa c=\dfrac{3\times10^{-5}}{10^{-2}}=3\times10^{-3} S m2^2 mol−1^{-1}.

Answer: 3

Topics:Conductance and Kohlrausch's lawSolubility product
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