Question

Calculate the equivalent conductivity of $1M$ aqueous diprotic acid solution if its conductivity is $50\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$.

• A. ${\Lambda }_{eq}=2.5\times {10}^{3}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• B. ${\Lambda }_{eq}=5\times {10}^{4}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• C. ${\Lambda }_{eq}=2.5\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$
• D. ${\Lambda }_{eq}=5\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$

Answer 1

The correct option is C ${\Lambda }_{eq}=2.5\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$

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

As given solution is a diprotic acid
$n=2$

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


$\text{Specific conductivity,}\kappa =50\times {10}^{-2}oh{m}^{-1}c{m}^{-1}$

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

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

${\Lambda }_{eq}=\frac{1000\times 50\times {10}^{-2}}{2}$

${\Lambda }_{eq}=2.5\times {10}^{2}oh{m}^{-1}c{m}^{2}equi{v}^{-1}$