The molar conductivity of 0.02 M KCl solution is $410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ at…

The molar conductivity of 0.02 M KCl solution is $410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ at $25^{\circ} \mathrm{C}$. Calculate its conductivity?
  1. $8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  2. $2.8 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  3. $4.1 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  4. $5.4 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

$\begin{aligned} \Lambda_{\mathrm{m}} & =\frac{1000 \mathrm{k}}{\mathrm{c}} \\ \mathrm{k} & =\frac{\Lambda_{\mathrm{m}} \times \mathrm{c}}{1000}=\frac{410 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.02 \mathrm{~mol} \mathrm{dm}^{-3}}{1000 \mathrm{~cm}^3 \mathrm{dm}^{-3}} \\ & =8.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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