The conductivity of a solution containing $2.08 \mathrm{~g}$ of anhydrous barium chloride in $200…

The conductivity of a solution containing $2.08 \mathrm{~g}$ of anhydrous barium chloride in $200 \mathrm{~mL}$ solution is $6 \times 10^{-3}$ $\mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$. The molar conductivity of the solution (in $\left.\mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}\right)$ is $\underline{x} \times 10^2$. The value of $\underline{x}$ is (Atomic mass of $\mathrm{Ba}=137, \mathrm{Cl}=35.5$ )
  1. 1.2
  2. 2.4
  3. 3.6
  4. 3.0

Solution

Conductivity (k) $=6 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$ $ \mathrm{V}=200 \mathrm{~mL}=0.2 \mathrm{~L} $ mass $=2.08 \mathrm{~g}$ $ \begin{aligned} & \Rightarrow \mathrm{C}=\frac{\mathrm{n}}{\mathrm{V}}=\frac{\left(\frac{\text { mass }}{\text { molar mass }}\right)}{\mathrm{V}}=\frac{\left(\frac{2.08}{208}\right)}{0.2} \\ & =0.05 \mathrm{M} . \end{aligned} $ $ \begin{aligned} & \Rightarrow \Lambda_{\mathrm{m}}=\frac{\mathrm{K} \times 1000}{\mathrm{c}}=\frac{\left(6 \times 10^{-3}\right)(1000)}{0.05} \\ & =120=1.20 \times 10^2 \end{aligned} $ Thus, $\mathrm{x}=1.2$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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