Question

The specific conductivity of $0.1$N KCl solution is $0.0129oh{m}^{-1}$ $c{m}^{-1}$. The resistance of the solution in the cell is $100\Omega$. The cell constant of the cell will be.

• A. $1.10$
• B. $1.29$
• C. $0.56$
• D. $2.80$

Answer 1

The correct option is B $1.29$

The specific conductivity of $0.1$N KCl solution is $0.0129oh{m}^{-1}$ $c{m}^{-1}$. The resistance of the solution in the cell is $100\Omega$.
The specific conductivity $\left(k\right)=\frac{1}{R}\times$ cell constant

The cell constant $=k\times R$

The cell constant $=0.0129oh{m}^{-1}$ $c{m}^{-1}\times 100\Omega =1.29c{m}^{-1}$.
The cell constant of the cell will be $1.29c{m}^{-1}$.

Hence, option $\text{B}$ is correct