Question

\(0.1\) M solution of an electrolyte \(\text{A}^{+}\text{B}^{-}\) placed in a conductivity cell with electrodes \(4\) cm apart and each with cross-section area equal to \(2\ \text{cm}^{2}\) was found to have a resistance of \(200\ \Omega\). What is molar conductivity of the solution?

• A. \(100\ \Omega ^{-1}\text{cm}^{2}\ \text{mol}^{-1}\)
• B. \(250\ \Omega ^{-1}\text{cm}^{2}\ \text{mol}^{-1}\)
• C. \(0.250\ \Omega ^{-1}\text{cm}^{2}\ \text{mol}^{-1}\)
• D. \(400\ \Omega ^{-1}\text{cm}^{2}\ \text{mol}^{-1}\)

Hint:

Find the cell constant, then conductivity from resistance, and convert to molar conductivity using the concentration.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Cell constant \(=\dfrac{l}{A}=\dfrac{4}{2}=2\ \text{cm}^{-1}\). Conductivity \(\kappa=\dfrac{\text{cell constant}}{R}\).

Step 2: Conductivity
\(\kappa=\dfrac{2}{200}=0.01\ \Omega^{-1}\text{cm}^{-1}\).

Step 3: Calculation
\(\Lambda_{m}=\dfrac{\kappa\times1000}{C}=\dfrac{0.01\times1000}{0.1}=100\ \Omega^{-1}\text{cm}^{2}\text{mol}^{-1}\).

Final Answer:
The molar conductivity is 100 ohm inverse cm squared per mol. \[ \boxed{\text{(A) }100\ \Omega^{-1}\text{cm}^{2}\ \text{mol}^{-1}} \]