Calculate the equivalent conductivity of $1 \mathrm{M} {\mathrm{H}}_2{\mathrm{SO}}_4$solution, if its conductivity is $26 \mathrm{x} {10}^{- 2} {\mathrm{ohm}}^{-} {\mathrm{cm}}^{-}$(at. wt. of sulphur $= 32)$

Molarity of ${\mathrm{H}}_2{\mathrm{SO}}_4=1\mathrm{M};$

$\text{ mol. wt. of }{\mathrm{H}}_2{\mathrm{SO}}_4=(2\times 1)+32+(4\times 16)=98 \mathrm{g} {\mathrm{mol}}^{-};$

Eq. wt. of ${\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\text{ Mol. wt. }}{2}=\frac{98}{2}=49{\text{ g. equiv}}^{\text{-}}$

But Normality x Eq. Wt. = Molarity x mol. wt.

$\therefore$ Normality of 

$\begin{array}{l}{\mathrm{H}}_2{\mathrm{SO}}_4=\frac{\text{ Molarity }\times \text{ mol. wt. }}{\text{ Eq. wt. }}\\ =\frac{1 \mathrm{mol} {\mathrm{L}}^{-}\times 98 \mathrm{g} {\mathrm{mol}}^{-}}{49 \mathrm{g}{\text{. equiv }}^{-}}\\ =2 \mathrm{eq}.{\mathrm{L}}^{-}\end{array}$

$\therefore$ Equivalent conductivity,

$\begin{array}{l}\wedge =\frac{\mathrm{K}\times \frac{1000{\mathrm{cm}}^3}{\mathrm{L}}}{\text{ Normality }}\\ =\frac{26\times {10}^{-2}{\mathrm{ohm}}^{-}{\mathrm{cm}}^{-}\times 1000 {\mathrm{cm}}^3 {\mathrm{L}}^{-}}{2\text{ g. equiv }{\mathrm{L}}^{-}}\\ =130{\text{ ohm }}^{-}{\mathrm{cm}}^2{\text{ (g. equiv) }}^{-}\text{Ans. }\end{array}$