Question

The ratio of equivalent conductance to specific conductance of a $0.04NKCl$ solution at a particular temperature is:

• A. $20\times {10}^{4}c{m}^{3}equi{v}^{-1}$
• B. $25\times {10}^{3}c{m}^{3}equi{v}^{-1}$
• C. $25\times {10}^{4}c{m}^{3}equi{v}^{-1}$
• D. $25\times {10}^{5}c{m}^{3}equi{v}^{-1}$

Answer 1

The correct option is B $25\times {10}^{3}c{m}^{3}equi{v}^{-1}$

Equivalent conductance is the conductivity of a solution containing 1 g-equivalent of the electrolyte.

${\Lambda }_{eq}=\kappa V$
$V$ is volume of solution containing $1g-equiv$

$\text{Normality of solution}\to N$
$Ng-equiv\text{of solute present in}\to 1L\text{solution}$

$1g-equiv\text{of solute present in}\to \frac{1}{N}L\text{solution}$
$\therefore$
${\Lambda }_{eq}=\frac{\kappa }{N}\left(\text{when volume in litres}\right)$

${\Lambda }_{eq}=\frac{\kappa \times 1000}{N}\left(\text{when volume in}c{m}^3\right)$

$\frac{{\Lambda }_{eq}}{\kappa }=\frac{1000}{N}$

$\frac{{\Lambda }_{eq}}{\kappa }=\frac{1000}{0.04}=25\times {10}^{3}c{m}^{3}equi{v}^{-1}$

At a particular temperature, the ratio of equivalent conductance to specific conductance of a $0.04KCl$ solution is $25\times {10}^{3}c{m}^{3}equi{v}^{-1}$