Chemistry - Electrochemistry Question with Solution | TestHub

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
$\begin{array}{rl}& {\lambda }_{\mathrm{m}}=\frac{\mathrm{K}}{\mathrm{M}}=\frac{\frac{1}{8.7}}{\frac{2}{40}}=\frac{40}{2\times 8.7}=2.29\mathrm{S}{\mathrm{dm}}^2{\mathrm{mol}}^{-1}\\ & 2.29\times 10^3\mathrm{m}\mathrm{S}{\mathrm{dm}}^2{\mathrm{mol}}^{-1}\\ & =22.9\times 10^2\mathrm{m}\mathrm{S}{\mathrm{dm}}^2{\mathrm{mol}}^{-1}\\ & =23\times 10^2\mathrm{m}\mathrm{S}{\mathrm{dm}}^2{\mathrm{mol}}^{-1}\end{array}$