The molar conductivity of $0.4 \mathrm{M} \mathrm{KCl}$ solution is $2.5 \times 10^5 \Omega^{-1}$…

The molar conductivity of $0.4 \mathrm{M} \mathrm{KCl}$ solution is $2.5 \times 10^5 \Omega^{-1}$ $\mathrm{cm}^2 \mathrm{~mol}^{-1}$. What is the resistivity of solution?
  1. $2.1 \times 10^2$
  2. $2.5 \times 10^2$
  3. $1 \times 10^{-2}$
  4. $2.8 \times 10^{-2}$

Solution

$\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 1000}{\mathrm{M}} \\ & 2.5 \times 10^5=\frac{\mathrm{k} \times 1000}{0.4} \\ & \text { Conductivity }(\mathrm{k})=\frac{1}{\rho(\text { resistivity })} \\ & =\frac{2.5 \times 10^5 \times 0.4}{1000} \\ & =100 \Omega^{-1} \mathrm{~cm}^{-1} \\ & \rho=\frac{1}{\mathrm{k}}=\frac{1}{100}=10^{-2} \Omega \mathrm{cm} \end{aligned}$

Asked in: MHT CET 2021 (21 Sep Shift 1)

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