Given : Concentration of NaOH=0.2%(w/v)
∴0.2 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 m ohm m =870×10−3ohmm
=870×10−3×10ohmdm
=870×10−2 ohm dm
=8.7 ohm dm
Now conductivity
K=ρ1=8.71ohm−1dm−1
Now molar conductivity of solution is
$\begin{aligned}
& \lambda_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{M}}=\frac{\frac{1}{8.7}}{\frac{2}{40}}=\frac{40}{2 \times 8.7}=2.29 \mathrm{S} \mathrm{dm}^2 \mathrm{mol}^{-1} \
& 2.29 \times 10^3 \mathrm{m} \mathrm{S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \
& =22.9 \times 10^2 \mathrm{m} \mathrm{S} \mathrm{dm}^2 \mathrm{~mol}^{-1} \
& =23 \times 10^2 \mathrm{m} \mathrm{S} \mathrm{dm}^2 \mathrm{~mol}^{-1}
\end{aligned}$