The conductivity of 0.005 M NaI solution at $25^{\circ} \mathrm{C}$ is $6.07 \times 10^{-4} \Omega^{-1}…

The conductivity of 0.005 M NaI solution at $25^{\circ} \mathrm{C}$ is $6.07 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$. Calculate its molar conductivity
  1. $121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  2. $110.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  3. $201.1 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  4. $241.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} \Lambda_{\mathrm{m}} & =\frac{1000 \mathrm{k}}{\mathrm{c}}=\frac{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1} \times 6.07 \times 10^{-4} \Omega^{-1} \cdot \mathrm{~cm}^{-1}}{0.005 \mathrm{~mol} \mathrm{~L}^{-1}} \\ & =121.4 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 1)

Practice more Electrochemistry questions on Aicharya