Ksp(A2X3)=1.1×10−23=110×10−25
Ksp=(2)2(3)3(s)5=110×10−25
4×27(s)5=110×10−25;S=M=1×10−5
K=3×10−5Sm−1
λm=[x]×10−3Sm2mol−1
λm=[MK×10−3]
=1×10−53×10−5×10−3=3×10−3Sm2mol−1
The solubility product of a sparingly soluble salt A_2 X_3 is 1.1× 10^-23. If specific conductance of the solution is 3× 10^-5 Sm^-1, the limiting…
JEE Main 2022 — Chemistry Physical Chemistry
2022integerhard
The solubility product of a sparingly soluble salt A2X3 is 1.1×10−23. If specific conductance of the solution is 3×10−5Sm−1, the limiting molar conductivity of the solution is x×10−3Sm2mol−1. The value of x is
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Held on 28 Jun 2022 · Verified 6 Jul 2026.
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