Question

Calculate the equivalent conductivity of $1\text{M}{H}_{2}S{O}_4$ solution if its specific conductivity is $26\times {10}^{-2}{\text{ohm}}^{-1}{\text{cm}}^{-1}$.

$(\text{Atomic weight of sulphur}=32\text{g/mol})$

Answer 1

Relation betweem molarity and normality:
$\text{Normality}=n\times \text{Molarity}$
$n$ is $n$-factor

As $H_{2}S{O}_4$ is a diprotic acid
so, $n=2$

Thus,
$N=nM=2\times 1M=2N$


$\text{Specific conductivity,}\kappa =26\times {10}^{-2}{\text{ohm}}^{-1}{\text{cm}}^{-1}$

$1\text{L}=1000{\text{cm}}^3$
$1{\text{L}}^{-1}={10}^{-3}{\text{cm}}^{-3}$
$\text{Concentration,}\left(C\right)=2{\text{g equiv L}}^{-1}$
$\text{Concentration,}\left(C\right)=2\times {10}^{-3}{\text{g equiv cm}}^{-3}$

If $\kappa$ is expressed in ${\text{S cm}}^{-1}$ and $C$ in ${\text{g equiv cm}}^{-3}$
$\Rightarrow {\Lambda }_{eq}=\frac{\kappa \times 1000}{\text{N}}$

$\Rightarrow {\Lambda }_{eq}=\frac{1000\times 26\times {10}^{-2}}{2}$

$\Rightarrow {\Lambda }_{eq}=1.3\times {10}^2{\text{ohm}}^{-1}{\text{cm}}^2{\text{equiv}}^{-1}$