Question

Specific conductance of $N/50KCl$ solution at ${25}^{\circ }C$ is $0.002765mhoc{m}^{-1}$. If the resistance of cell along with the solution is $400ohm$, then cell constant will be:

• A. $0.553c{m}^{-1}$
• B. $1.106c{m}^{-1}$
• C. $2.212c{m}^{-1}$
• D. None of these

Answer 1

Answer: (B)

$1.106c{m}^{-1}$

Specific conductance is $0.002765mhoc{m}^{-1}$.
The resistance is $400ohm$,

The cell constant

$\frac{l}{a}=\kappa \times R=0.002765mhoc{m}^{-1}\times 400ohm=1.106c{m}^{-1}$