Resistance of the conductivity of a cell filled with 0.1 M KCl solution is 100 $Ω$ . If the resistance of the same cell when filled with the 0.02 M KCl solution is 520 $Ω$ . Then the molar conductivity of 0.02 M KCl solution will be [The conductivity of 0.1 M KCl solution is 1.29 S.m^–1]

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124 S cm² mole^–1

1.24 S.m² mole^–1

124 S.m² mole^–1

1.24 S.cm² mole^–1

answer is C.

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Detailed Solution

For 0.1 M KCl
$\begin{array}{l} K = \frac{1}{R} G^{x} 1.29 = \frac{G^{x}}{100} \\ G^{x} = 129 m^{- 1} or 1.29 {cm}^{- 1} \\ Λ_{M} = \frac{1000 \times k}{M} = \frac{1000 \times 1.29}{520 \times 2 \times 10^{- 2}} \end{array}$
=124 S.cm² mole^–1

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