Question

Resistance of $0.2N$ solution of an electrolyte is $40\Omega$. The specific conductance of the solution is $1.4S{m}^{-1}$. Find the equivalent conductivity of the given solution.

• A. ${\Lambda }_{eq}=1.5\times {10}^{-4}S{m}^{2}e{q}^{-1}$
• B. ${\Lambda }_{eq}=3.5\times {10}^{-4}S{m}^{2}e{q}^{-1}$
• C. ${\Lambda }_{eq}=7\times {10}^{-3}S{m}^{2}e{q}^{-1}$
• D. ${\Lambda }_{eq}=2\times {10}^{-3}S{m}^{2}e{q}^{-1}$

Answer 1

The correct option is C ${\Lambda }_{eq}=7\times {10}^{-3}S{m}^{2}e{q}^{-1}$

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 $1g-equiv$

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

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

If $\kappa$ is expressed in $S{m}^{-1}$
then,
${\Lambda }_{eq}\left(S{m}^{2}e{q}^{-1}\right)=\frac{\kappa \left(S{m}^{-1}\right)}{1000L{m}^{-3}\times \text{Normality}\left(eq{L}^{-1}\right)}$

${\Lambda }_{eq}\left(S{m}^{2}e{q}^{-1}\right)=\frac{\kappa \left(S{m}^{-1}\right)}{1000L{m}^{-3}\times \text{Normality}\left(eq{L}^{-1}\right)}$

${\Lambda }_{eq}=\frac{1.4}{1000\times 0.2}$

${\Lambda }_{eq}=7\times {10}^{-3}S{m}^{2}e{q}^{-1}$