Question

The specific conductivity of $0.1NKCI$ 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

Answer: (B)

The specific conductivity of $0.1NKCl$ solution is $0.0129oh{m}^{-1}c{m}^{-1}$.
The resistance of the solution in the cell is $100\Omega$.
The specific conductivity $(k)=\frac{1}{R}\times$ 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 $1.29c{m}^{-1}$.