Question

The specific conductivity of $0.1\, N\, KCI$ solution is $0.0129\, ohm^{-1}$ $cm^{-1}$. 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 $0.1\, N \,KCl$ solution is $0.0129\, ohm ^{-1} \,cm ^{-1}$.
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 $=0.0129\, ohm ^{-1}\, cm ^{-1} \times 100\, \Omega=1.29\, cm ^{-1}$.
The cell constant of the cell will be $1.29 \,cm ^{-1}$.