Q12 / 120 · VITEEE 2014 · Chemistry

The equivalent conductivity of a solution containing $2.54 \mathrm{~g}$ of $\mathrm{CuSO}_{4}$ per $\mathrm{L}$ is $91.0 \mathrm{~W}^{-1}$ $\mathrm{cm}^{2} \mathrm{eq}^{-1}$. Its conductivity would be

  • Correct answer
    A
    $2.9 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
  • B
    $1.8 \times 10^{-2} \Omega^{-1} \mathrm{~cm}^{-1}$
  • C
    $2.4 \times 10^{-4} \Omega^{-1} \mathrm{~cm}^{-1}$
  • D
    $3.6 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$

Solution

We know that, $\begin{array}{l} \kappa=\Lambda_{\mathrm{eq}} \cdot \mathrm{C} \\ \text { Given, } \Lambda \mathrm{eq}=91.0 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1} \\ \kappa=\left(91 \Omega^{-1} \mathrm{~cm}^{2} \mathrm{eq}^{-1}\right) \\ =\left(\frac{2.54}{159 / 2 \times 1000} \mathrm{eq} \mathrm{cm}^{-3}\right) \\ =2.9 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1} \end{array}$

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

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