Question

Resistance of a conductivity cell filled with \(0.1\, \text{mol L}^{-1}\) NaCl is 100 \(\Omega\).
Resistance of the same cell with \(0.02\, \text{mol L}^{-1}\) NaCl = 258 \(\Omega\). Conductivity of \(0.1\, \text{mol L}^{-1}\) NaCl = \(1.29\, \text{S m}^{-1}\). What is conductivity of 0.02 M NaCl?

• A. \(1.0\, \text{S m}^{-1}\)
• B. \(0.2\, \text{S m}^{-1}\)
• C. \(2.0\, \text{S m}^{-1}\)
• D. \(0.5\, \text{S m}^{-1}\)

Hint:

Use \( \kappa = \frac{\text{cell constant}}{R} \). If cell constant is unknown, use known data to compute it first.

Answer 1

The Correct Option is D

Use: \[ \kappa = \frac{\text{Cell Constant}}{R} \] Let cell constant = \( K \Rightarrow \frac{1.29}{100} = \frac{K}{100} \Rightarrow K = 129 \) Now, \[ \kappa_{0.02M} = \frac{129}{258} = 0.5 \, \text{S m}^{-1} \]