Question

In a conductivity cell the two platinum electrodes, each of area $10{\text{cm}}^2,$ are fixed $1.5\text{cm}$ apart. The cell contains $0.05\text{N}$ solution of a salt. If the two electrodes are just half dipped into the solution which has a resistance of $50\Omega$, find specific conductance of the salt solution.

• A. $0.006{\Omega }^{-1}{\text{cm}}^{-1}$
• B. $0.004{\Omega }^{-1}{\text{cm}}^{-1}$
• C. $0.003{\Omega }^{-1}{\text{cm}}^{-1}$
• D. $0.005{\Omega }^{-1}{\text{cm}}^{-1}$

Answer 1

The correct option is A $0.006{\Omega }^{-1}{\text{cm}}^{-1}$

Since, the electrodes of the cell are just half dipped, the effective area$\left(a\right)$ will be $5{\text{cm}}^2$.
$l=1.5\text{cm}$
$\text{Cell constant}=\frac{l}{a}=\frac{1.5}{5}=0.3{\text{cm}}^{-1}$
$\text{Specific conductance}=\text{conductance}\times \text{cell constant}=\frac{1}{\text{resistance}}\times \text{cell constant}=\frac{1}{50}\times 0.3=0.006{\Omega }^{-1}{\text{cm}}^{-1}$