Question

The resistance of a $0.02N$ solution of an electrolyte $MgC{l}_2$ was found to be $210ohm$ at $298K$ using a conductivity cell with a cell constant of $0.88c{m}^{-1}$. Calculate the equivalent conductivity of the solution.

• A. $625oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• B. $500oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• C. $209.5oh{m}^{-1}c{m}^{2}e{q}^{-1}$
• D. $125.5oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Answer 1

The correct option is C $209.5oh{m}^{-1}c{m}^{2}e{q}^{-1}$

Given:
$\text{Cell constant,}\frac{l}{A}=0.88c{m}^{-1}\text{Resistance,}R=210ohm$

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

$\kappa =0.88\times \frac{1}{210}=4.19\times {10}^{-3}ohm{}^{-1}c{m}^{-1}$

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 $1g-equiv$

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

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

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

${\Lambda }_{eq}=\frac{4.19\times {10}^{-3}\times 1000}{0.02}$


${\Lambda }_{eq}=209.5oh{m}^{-1}c{m}^{2}e{q}^{-1}$