The conductivity of 0.20 {\M} solution of {\KCl} at 298 {\K} is 0.025 {\S} {\cm}^{-1}, Calculate its molar conductivity.

CBSE Class 12 Chemistry 2013 Solved Paper - Question 10

Question : 10 of 30

Marks: +1, -0

The conductivity of

0.20 {\M}

solution of

{\KCl}

298 {\K}

0.025 {\S} {\cm}^{-1}

, Calculate its molar conductivity.

Solution:

\;\text " Conductivity "\;\;=0.025 {\S} {\cm}^{-1}

\;\text " molarity, "\; {\M}\;=0.20 {\M}

{T} =298 { K}

{T}_{{m}} =?

molar conductivity

=?

{T}_{{m}}= {\κ\×1000}/{{C}}

={0.025 \×1000 {S} {cm}^{-1}}/{0.20 {\mol} {cm}^{-3}}

={25}/{0.20}

=125 \ {\S}\ {\cm}^2 {\mol}^{-1}

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