Question

What is the conductivity of \(0.02 \text{M} \text{AgNO}_3\) solution having cell constant \(1.2 \text{cm}^{-1}\) and resistance \(95.0 \text{ohms}\) ?

• A. \(1.140\times 10^{-2} \Omega ^{-1} \text{cm}^{-1}\)
• B. \(1.263\times 10^{-2} \Omega ^{-1} \text{cm}^{-1}\)
• C. \(1.340\times 10^{-2} \Omega ^{-1} \text{cm}^{-1}\)
• D. \(1.463\times 10^{-2} \Omega ^{-1} \text{cm}^{-1}\)

Hint:

Conductivity is the cell constant divided by resistance.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The conductivity \(\kappa\) of a solution is the cell constant \(G^*\) times the conductance, and conductance is \(1/R\).

Step 2: Key Formula:
\[ \kappa = \frac{G^*}{R} \]

Step 3: Calculation:
\[ \kappa = \frac{1.2\ \text{cm}^{-1}}{95.0\ \Omega} = 1.263\times 10^{-2}\ \Omega^{-1}\text{cm}^{-1} \]

Step 4: Check the Other Options:
Multiplying instead of dividing would give 114, a huge number. The other options (1.140, 1.340, 1.463 times \(10^{-2}\)) correspond to R values of 105, 89.6 and 82, none of them 95.0 ohm. So (B) is correct.

Final Answer:
The conductivity is \(1.263\times10^{-2}\ \Omega^{-1}\text{cm}^{-1}\), option (B). \[ \boxed{\text{(B) } 1.263\times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}} \]