Question

The measured resistance of a conductance cell containing $7.5\times {10}^{-3}M$ solution of $KCl$ at ${25}^{o}C$ was $1005ohms$. Calculate molar conductance $\left(oh{m}^{-1}c{m}^{2}mo{l}^{-1}\right)$ of the solution.
(Cell constant$=1.25c{m}^{-1}$)

• A. $16.587$
• B. $165.87$
• C. $1658.7$
• D. $1.6587$

Answer 1

The correct option is B $165.87$

As conductance= $\frac{1}{Resistance}=K\frac{A}{l}$

where, $K$= specific conductivity

$l/A$= cell constant

$\Rightarrow \frac{1}{1005ohms}=K\frac{1}{1.25c{m}^{-1}}$

$K=1.243\times {10}^{-3}oh{m}^{-1}c{m}^{-1}$

as molar conductivity= $\frac{\text{specific conductivity}}{concentration}$

$\Rightarrow {\wedge }_m=\frac{1.243\times {10}^{-3}oh{m}^{-1}c{n}^{-1}}{7.5\times {10}^{-3}mol{L}^{-1}}$

$\Rightarrow {\wedge }_m=0.16587oh{m}^{-1}c{m}^{-1}mo{l}^{-1}L$

$\Rightarrow {\wedge }_m=0.16587oh{m}^{-1}c{m}^{-1}mo{l}^{-1}\times 1000ml$

$\Rightarrow {\wedge }_m=165.87oh{m}^{-1}c{m}^{-1}mo{l}^{-1}c{m}^3$ as $1ml=1c{m}^3$

$\Rightarrow {\wedge }_m=165.87oh{m}^{-1}c{m}^{2}mo{l}^{-1}$