$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be ______ $ imes 10^2 \mathrm{mS} \mathrm{dm}{ }^2 \mathrm{~mol}^{-1}$. (Nearest integer)

Question

$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega \mathrm{~m}$. The molar conductivity of the solution will be ______ $	imes 10^2 \mathrm{mS} \mathrm{dm}{ }^2 \mathrm{~mol}^{-1}$. (Nearest integer)

Aicharya · Accepted Answer

Given : Concentration of $\mathrm{NaOH}=0.2 \%(\mathrm{w} / \mathrm{v})$
$\therefore 0.2 \mathrm{~g}$ of NaOH in 100 ml of solution.
Molarity of NaOH solution
$\begin{array}{r}
=\frac{\text { moles of solute }}{\mathrm{V}_{\mathrm{ml}}} \times 1000 \\
=\frac{0.2 / 40}{100} \times 1000==\frac{0.2}{40 \times 100} \times 1000=\frac{2}{40} \mathrm{M}
\end{array}$
Given resistivity of solution $=870 \mathrm{~m}$ ohm m $=870 \times 10^{-3} \mathrm{ohm} \mathrm{m}$
$=870 \times 10^{-3} \times 10 \mathrm{ohm} \mathrm{dm}$
$=870 \times 10^{-2}$ ohm dm
$=8.7$ ohm dm
Now conductivity
$\mathrm{K}=\frac{1}{\rho}=\frac{1}{8.7} \mathrm{ohm}^{-1} \mathrm{dm}^{-1}$
Now molar conductivity of solution is
$\begin{aligned}
& \lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{M}}=\frac{\frac{1}{8.7}}{\frac{2}{40}}=\frac{40}{2 \times 8.7}=2.29 \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\
& 2.29 \times 10^3 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\
& =22.9 \times 10^2 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \\
& =23 \times 10^2 \mathrm{~m} \mathrm{~S} \mathrm{dm}^2 \mathrm{~mol}^{-1}
\end{aligned}$

$0.2 \%(\mathrm{w} / \mathrm{v})$ solution of NaOH is measured to have resistivity $870.0 \mathrm{~m} \Omega…

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