Question

The resistance of $0.5M$ solution of an electrolyte in a cell was found to be $50\Omega$. If the electrodes in the cell are $2.2cm$ apart $\&$ having an area of $4.4c{m}^2$ then the molar conductivity (in $S{m}^{2}mo{l}^{-1}$ ) of solution is:

• A. 0.2
• B. 0.02
• C. 0.002
• D. None of these

Answer 1

Answer: (C)

Apply, $K=\frac{1}{R}\frac{\ell }{a}$

$\Rightarrow 50\Omega =\frac{1}{K}\times \frac{2.2\times 10^{-2}m}{4.4\times {\left(10^{-2}m\right)}^2}$

$K=1{\Omega }^{-1}{m}^{-1}$ or

Then, use ${\Lambda }_m=K\times v(ml)\to$ this is the volume which

contains 1 mole of electrolyte

${\Lambda }_m=1m^{-1}\times 2000c{m}^{3}mo{l}^{-1}$

${\Lambda }_m=1m^{-1}\times 2000\times 10^{-6}{m}^{3}mo{l}^{-1}(0.5$ moles are in $11tr$

so 1 mole is present in $2L$ )

${\Lambda }_m=2\times 10^{-3}S{m}^{2}mo{l}^{-1}$