If resistivity of $0.8 \mathrm{M} \mathrm{KCl}$ solution is $2 \cdot 5 \times 10^{-3} \Omega \mathrm{cm} .$…

If resistivity of $0.8 \mathrm{M} \mathrm{KCl}$ solution is $2 \cdot 5 \times 10^{-3} \Omega \mathrm{cm} .$ Calculate molar conductivity of solution?
  1. $3 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  2. $2 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  3. $4 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$
  4. $5 \times 10^{5} \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$

Solution

$\begin{array}{l} \text { Conductivity }(\mathrm{k})=\frac{1}{\text { resistivity }}=\frac{1}{2.5 \times 10^{-3} \Omega \mathrm{cm}}=0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1} \\ \text { Molar conductivity } \quad\left(\Lambda_{\mathrm{m}}\right)=\frac{1000 \mathrm{k}}{\mathrm{C}}=\frac{1000 \mathrm{~cm}^{3} \mathrm{~L}^{-1} \times 0.4 \times 10^{3} \Omega^{-1} \mathrm{~cm}^{-1}}{0.8 \mathrm{~mol} \mathrm{~L}^{-1}} \end{array}$

Asked in: MHT CET 2020 (16 Oct Shift 2)

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