Question

Resistance of a conductivity cell filled with \(0.1 \text{ mol L}^{-1}\) NaCl is \(100 \, \Omega\). If the resistance of the same cell when filled with \(0.02 \text{ mol L}^{-1}\) NaCl solution is \(258 \, \Omega\), then the conductivity of \(0.02 \text{ mol L}^{-1}\) NaCl solution is:
\[ \text{(Conductivity of }0.1 \text{ mol L}^{-1}\text{ NaCl} = 1.29 \, \text{S m}^{-1}) \]

• 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:

For the same conductivity cell, cell constant remains unchanged, so: \[ \kappa R = \text{constant} \] Conductivity is inversely proportional to resistance.

Answer 1

The Correct Option is D

Step 1: Use the relation between conductivity and resistance.
For a conductivity cell: :contentReference[oaicite:0]index=0 For the same cell, cell constant remains constant.
Therefore, \[ \kappa_1 R_1 = \kappa_2 R_2 \]

Step 2: Write the given values.
For \(0.1 \text{ mol L}^{-1}\) NaCl: \[ \kappa_1 = 1.29 \, \text{S m}^{-1} \] \[ R_1 = 100 \, \Omega \] For \(0.02 \text{ mol L}^{-1}\) NaCl: \[ R_2 = 258 \, \Omega \] Let conductivity be \(\kappa_2\).

Step 3: Apply the formula.
\[ \kappa_1 R_1 = \kappa_2 R_2 \] \[ 1.29 \times 100 = \kappa_2 \times 258 \] \[ 129 = 258\kappa_2 \]

Step 4: Calculate conductivity.
\[ \kappa_2 = \frac{129}{258} \] \[ \kappa_2 = 0.5 \, \text{S m}^{-1} \]

Step 5: Final Answer.
Hence, the conductivity of \(0.02 \text{ mol L}^{-1}\) NaCl solution is: \[ \boxed{0.5 \, \text{S m}^{-1}} \]