Question

The resistance of $0.01 \,m \,KCl$ solution at $298\, K$ is $1500\, \Omega$. If the conductivity of $0.01\, m \,KCl$ solution at $298\, K$ is $0.146 \,\times \,10^{-3}\, S \,cm ^{-1} .$ The cell constant of the conductivity cell in $cm ^{-1}$ is

• A. 0.219
• B. 0.291
• C. 0.301
• D. 0.194

Answer 1

The Correct Option is A

To find the cell constant of the conductivity cell, we need to use the relation between conductivity (\( \kappa \)), cell constant (\( G^* \)), and resistance (\( R \)). The formula is given by:

\[ \kappa = G^* \cdot \frac{1}{R} \]

Where:

• \( \kappa \) is the conductivity of the solution.
• \( G^* \) is the cell constant.
• \( R \) is the resistance.

We are given:

• Conductivity \( \kappa = 0.146 \times 10^{-3} \, S \,cm^{-1} \)
• Resistance \( R = 1500 \, \Omega \)

We can rearrange the formula to solve for the cell constant:

\[ G^* = \kappa \cdot R \]

Substitute the given values into the equation:

\[ G^* = 0.146 \times 10^{-3} \, S \,cm^{-1} \times 1500 \, \Omega \]

Perform the calculations:

\[ G^* = 0.146 \times 10^{-3} \times 1500 = 0.219 \, cm^{-1} \]

Thus, the cell constant of the conductivity cell is 0.219 cm^-1.

Hence, the correct answer is 0.219.