The conductivity of $0.012 \mathrm{M} \mathrm{NaBr}$ solution is $2.67 \times 10^{-4} \mathrm{~S}…

The conductivity of $0.012 \mathrm{M} \mathrm{NaBr}$ solution is $2.67 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1}$. What is it's molar conductivity?
  1. $26.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  2. $32.04 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  3. $12.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  4. $22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \quad \mathrm{K}=2.67 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \\ & \mathrm{M}=0.012 \mathrm{M} \\ & \Lambda_{\mathrm{m}}=\frac{2.67 \times 10^{-4} \times 1000}{0.012} \\ & =22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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