The molar conductivity of $0.1 \mathrm{M} \mathrm{BaCl}_2$ solution is $106 \Omega^{-1} \mathrm{~cm}^2$…

The molar conductivity of $0.1 \mathrm{M} \mathrm{BaCl}_2$ solution is $106 \Omega^{-1} \mathrm{~cm}^2$ $\mathrm{mol}^{-1}$ at $25^{\circ} \mathrm{C}$. What is it's conductivity?
  1. $1.06 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}$
  2. $5.03 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  3. $3.66 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  4. $2.6 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

$\wedge=\frac{1000 \mathrm{k}}{\mathrm{C}} \quad \therefore \mathrm{k}=\frac{\wedge \mathrm{c}}{1000}$ Now, $\wedge=106 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}, \mathrm{c}=0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ $\begin{aligned} & \therefore \mathrm{k}=\frac{106 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.1 \mathrm{molL}^{-1}}{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}} \\ & \therefore \mathrm{k}=1.06 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1} \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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