Q3 / 150 · MHT CET 2023 (09 May Shift 2) · Chemistry

What is the conductivity of $0.05 \mathrm{M} \mathrm{BaCl}_2$ solution if its molar conductivity is $220 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} ?$

  • Correct answer
    A
    $0.011 \Omega^{-1} \mathrm{~cm}^{-1}$
  • B
    $0.022 \Omega^{-1} \mathrm{~cm}^{-1}$
  • C
    $0.033 \Omega^{-1} \mathrm{~cm}^{-1}$
  • D
    $0.044 \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

$\begin{aligned} \Lambda & =\frac{1000 \mathrm{k}}{\mathrm{c}} \\ \mathrm{k} & =\frac{\Lambda \mathrm{c}}{1000} \\ \mathrm{k} & =\frac{220 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \times 0.05 \mathrm{~mol} \mathrm{~L}^{-1}}{1000 \mathrm{~cm}^3 \mathrm{~L}^{-1}} \\ & =0.011 \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}$

Hence, the correct answer is option: 'A'.

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