Q32 / 150 · MHT CET 2021 (24 Sep Shift 1) · Chemistry

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?

  • A
    $26.7 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • B
    $32.04 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • C
    $12.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
  • Correct answer
    D
    $22.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$

Solution

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

Hence, the correct answer is option: 'D'.

Relevant Questions