$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be__________$\times 10^2 \mathrm{mS} \mathrm{dm}^2 \mathrm{~mol}^{-1}$. (Nearest integer)
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Correct Answer: 23
Solution and Explanation
Conductivity $\kappa = \frac{1}{\rho} = \frac{1}{0.870\ \Omega\ m} = 1.149$ S m$^{-1}$.
$\Lambda_m = \frac{\kappa}{c} = \frac{1.149}{50} = 2.299 \times 10^{-2}$ S m$^2$ mol$^{-1}$.
1 S m$^2$ = $10^5$ mS dm$^2$, so $\Lambda_m = 2299$ mS dm$^2$ mol$^{-1} = 22.99 \times 10^2 \approx 23 \times 10^2$.
The answer is 23.
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