Question

Calculate the equivalent conductivity of $1M{H}_{2}S{O}_4$ solution if its conductivity is $26\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$.

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

• A. ${\Lambda }_{eq}=2.6\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• B. ${\Lambda }_{eq}=1.3\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• C. ${\Lambda }_{eq}=13\times {10}^{-5}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• D. ${\Lambda }_{eq}=26\times {10}^{-5}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$

Answer 1

The correct option is B ${\Lambda }_{eq}=1.3\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$

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


$\text{Specific conductivity,}\kappa =26\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$

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

If $\kappa$ is expressed in $Sc{m}^{-1}$ and $C$ in $g-equivc{m}^{-3}$
${\Lambda }_{eq}=\frac{\kappa \times 1000}{N}$

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

${\Lambda }_{eq}=1.3\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$