Question

The conductance of a salt solution $\left(AB\right)$ measured by two parallel electrodes of area $100{cm}^2$ separated by $10cm$ was found to be $0.0001{\Omega }^{-1}$. If volume enclosed between two electrode contain $0.1mole$ of salt, what is the molar conductivity of salt at same concentration?

• A. $10$
• B. $0.1$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

$10$

Cell constant= $l/A$

($l$= length, $A$= area)

For calculating molar conductivity, we have to calculate specific conductivity $\left(K\right)$ .

$K=G\times$ Cell constant ($G$= Conductance)

$\Rightarrow K=0.0001\times \frac{0.01}{0.01}=0.0001{\Omega }^{-1}{m}^{-1}$

Molar conductivity= $\frac{K\times 1000}{Molarity}$

Molarity= No. of moles/ Volume

$=0.01/1=0.01M$

Molar conductivity= $\frac{0.0001\times 1000}{0.01}$

$=10S{m}^{2}mo{l}^{-1}$