Molar conductivity of $0.01$ M HCl solution is $400 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$…

Molar conductivity of $0.01$ M HCl solution is $400 \cdot 0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. Calculate the conductivity of $\mathrm{HCl}$ solution.
  1. $4 \cdot 0 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
  2. $\cdot 8 \cdot 0 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}$
  3. $2 \cdot 5 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}$
  4. $4 \cdot 0 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

$\mathrm{C}=0.01 \mathrm{~M}, \quad \wedge=400.0 ~\Omega^{-1} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ $\mathrm{k}=?$ $\wedge=\frac{1000 \mathrm{~k}}{\mathrm{C}} \quad \therefore \mathrm{k}=\frac{\wedge \times \mathrm{C}}{1000}$ $\therefore \mathrm{k}=\frac{400.0 \times 0.01}{1000}=4.0 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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