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If the conductivity of a $0.5 \mathrm{M} \mathrm{KCl}$ solution at $298 \mathrm{~K}$ is $0.024 \mathrm{~S} \mathrm{~cm}^{-1}$, then the molar conductivity of the solution would be $\mathrm{Scm}^2 \mathrm{~mol}^{-1}$.
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The correct answer is:
48
Electrolytic conductivity, $\kappa=0.024 \mathrm{~S} \mathrm{~cm}^{-1}$
Molar concentration $=0.5 \mathrm{M}$
$$
\begin{aligned}
\Lambda_{\mathrm{m}}^0(\text { Molar conductivity }) & =\frac{\kappa \times 1000}{M} \\
=\frac{0.024 \times 1000}{0.5} & =48 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} .
\end{aligned}
$$
Molar concentration $=0.5 \mathrm{M}$
$$
\begin{aligned}
\Lambda_{\mathrm{m}}^0(\text { Molar conductivity }) & =\frac{\kappa \times 1000}{M} \\
=\frac{0.024 \times 1000}{0.5} & =48 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} .
\end{aligned}
$$
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