Question

What is the cell constant of \( \frac{N}{10} \) KCl solution at \(25^\circ\text{C}\), if conductivity and resistance of the solution are \(0.0112~\Omega^{-1}\text{cm}^{-1}\) and \(55.0~\Omega\) respectively?

• A. \(0.616~\text{cm}^{-1}\)
• B. \(0.491~\text{cm}^{-1}\)
• C. \(2.0~\text{cm}^{-1}\)
• D. \(0.2~\text{cm}^{-1}\)

Hint:

Always remember that the cell constant is the product of conductivity and resistance, not their ratio.

Answer 1

The Correct Option is A

Step 1: Write the formula for cell constant.
The relation between conductivity \( \kappa \), resistance \( R \), and cell constant \( G^\ast \) is given by: \[ G^\ast = \kappa \times R \] Step 2: Substitute the given values.
\[ \kappa = 0.0112~\Omega^{-1}\text{cm}^{-1}, \quad R = 55.0~\Omega \] \[ G^\ast = 0.0112 \times 55.0 \] Step 3: Calculate the result.
\[ G^\ast = 0.616~\text{cm}^{-1} \] Step 4: Final conclusion.
Hence, the cell constant of the given KCl solution is \(0.616~\text{cm}^{-1}\).