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JEE Main 202228 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
28 June
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2022 · 28 June · Morning Shift · Chemistry · Electrochemistry

The solubility product of a sparingly soluble salt A _ 2 X _ 3 is 1...

The solubility product of a sparingly soluble salt A2X3{A}_{2}{X}_{3} is 1.1×10−231 . 1 \times {10}^{- 23}. If specific conductance of the solution is 3×10−5Sm−13 \times {10}^{- 5}{Sm}^{- 1}, the limiting molar conductivity of the solution is x×10−3Sm2mol−1x \times {10}^{- 3}S{m}^{2}{mol}^{- 1}. The value of xx is

Correct answer

0

Step-by-step solution

Ksp(A2X3)=1.1×10−23=110×10−25{K}_{sp}\left({A}_{2}{X}_{3}\right) = 1 . 1 \times {10}^{- 23} = 110 \times {10}^{- 25}

Ksp=(2)2(3)3(s)5=110×10−25{K}_{sp} = {\left(2\right)}^{2}{\left(3\right)}^{3}{\left(s\right)}^{5} = 110 \times {10}^{- 25}

4×27(s)5=110×10−25;S=M=1×10−54 \times 27{\left(s\right)}^{5} = 110 \times {10}^{- 25}; S = M = 1 \times {10}^{- 5}

K=3×10−5Sm−1K = 3 \times 10 - 5{Sm}^{- 1}

λm=[x]×10−3Sm2mol−1{\lambda}_{m} = \left[x\right] \times {10}^{- 3}{Sm}^{2}{mol}^{- 1}

λm=[K×10−3M]{\lambda}_{m} = \left[\dfrac{K \times {10}^{- 3}}{M}\right]

=3×10−5×10−31×10−5=3×10−3Sm2mol−1= \dfrac{3 \times {10}^{- 5} \times {10}^{- 3}}{1 \times {10}^{- 5}} = 3 \times {10}^{- 3}{Sm}^{2}{mol}^{- 1}

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