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Molar conductivity of $0.04 \mathrm{M} \mathrm{BaCl}_2$ solution is $230 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ at $27^{\circ} \mathrm{C}$. what is its conductivity?
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The correct answer is:
$9.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}$
$\begin{aligned} & \Lambda_{\mathrm{m}}=\frac{\mathrm{k} \times 100}{\mathrm{M}} \\ & 230=\frac{\mathrm{k} \times 100}{0.04} \\ & \mathrm{k}=\frac{230 \times 0.04}{1000} \\ & =9.2 \times 10^{-3} \Omega^{-1} \mathrm{~cm}^{-1}\end{aligned}$
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