A cell whose resistance, when filled with 0.1 M – KCl is 200 Ω, is measured to be 6400 Ω, when fille

Question

A cell whose resistance, when filled with 0.1 M – KCl is 200 Ω, is measured to be 6400 Ω, when filled with 0.003 M – NaCl solution. What is the molar conductance of NaCl solution, in Ω−1 cm2 mol−1 if the molar conductance of 0.1 M – KCl is 120 Ω−1 cm2 mol−1?

Accepted Answer

$\frac{\Lambda_{m}(NaCl)}{\Lambda_{m}(KCl)} = \frac{(\kappa/C)_{NaCl}}{(\kappa/C)_{KCl}} = \frac{\frac{1}{R_{NaCl}}.G^{*}.\frac{1}{C_{NaCl}}}{\frac{1}{R_{KCl}}.G^{*}.\frac{1}{C_{KCl}}} = \frac{(R.C)_{KCl}}{(R.C)_{NaCl}}$

$\text{Or, }\frac{\Lambda_{m}(NaCl)}{120} = \frac{200 \times 0.1}{6400 \times 0.003} \Rightarrow \Lambda_{m(NaCl)} = 125\Omega^{- 1}\text{c}\text{m}^{- 1}\text{mo}\text{l}^{- 1}$

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