Question

$1.0N$ solution of a salt surrounding two platinum electrodes of $2.1cm$ apart and $4.2c{m}^2$ in area was found to offer a resistance of $50ohm$. Calculate the equivalent conductivity of the solution.

• A. ${\Lambda }_{eq}={10}^{-5}oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• B. ${\Lambda }_{eq}=0.01oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• C. ${\Lambda }_{eq}=100oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• D. ${\Lambda }_{eq}=10oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Answer 1

The correct option is D ${\Lambda }_{eq}=10oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Given:
$\text{Distance between electrodes,}l=2.1cm\text{Area,}A=4.2c{m}^2\text{Resistance,}R=50ohm$

$\text{Specific conductance,}\kappa =\frac{l}{A}\times \frac{1}{R}$

$\kappa =\frac{2.1}{4.2}\times \frac{1}{50}=0.01ohm{}^{-1}c{m}^{-1}$

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 $1g-equiv$

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

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

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

${\Lambda }_{eq}=\frac{0.01\times 1000}{1}$

${\Lambda }_{eq}=10oh{m}^{-1}c{m}^{2}e{q}^{-1}$