Question

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

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

Answer 1

The correct option is C 0.002

conductivity,$K=G^{\ast }\frac{1}{R}$
where $G^{\ast }$ is the cell constant $G^{\ast }=\frac{l}{A}$
where l is the distance between the electrodes and A is the area.

Therefore, $K=\frac{1}{50}\times \frac{2.2}{4.4}$

= $0.01{\Omega }^{–1}c{m}^{–1}$
molar conductivity, $\lambda m=K\times \frac{1000}{C}$

=$0.01\times \frac{1000}{0.5}$

= $20{\Omega }^{–1}c{m}^{2}mo{l}^{–1}$

=$0.002{\Omega }^{–1}{m}^{2}mo{l}^{–1}$