Question

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 ohm. The cell constant of the cell will be

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

Hint:

Cell constant = l/A, units cm\(^{-1}\).

Answer 1

The Correct Option is B

To determine the cell constant, we need to use the following relationship between specific conductivity (or conductivity), resistance, and the cell constant:

\(k = \frac{1}{R} \times K\)

Where:

• \(k\) is the conductivity.
• \(R\) is the resistance measured in ohms.
• \(K\) is the cell constant, which we need to find.

Given in the problem:

• Specific conductivity, \(k = 0.0129 \, \text{ohm}^{-1} \, \text{cm}^{-1}\)
• Resistance, \(R = 100 \, \text{ohms}\)

Substitute these values into the equation:

\(0.0129 = \frac{1}{100} \times K\)

Rearrange the equation to solve for \(K\):

\(K = 0.0129 \times 100\)

Calculate the value of \(K\):

\(K = 1.29\)

Therefore, the cell constant of the cell is 1.29.

Conclusion: The correct answer is 1.29, so option 2 is the correct choice.