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: Method:
Find the conductance first: $G = 1/95 = 0.010526\ \Omega^{-1}$.

Step 2: Multiply by Cell Constant:
$\kappa = 0.010526 \times 1.2 = 0.012632$.

Step 3: Answer:
This is $1.263\times10^{-2}\ \Omega^{-1}\text{cm}^{-1}$, option (B).

Final Answer:
Option (B). \[ \boxed{\text{(B) } 1.263\times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}} \]