Question

Calculate the conductivity of \(0.025\) M \(\text{BaCl}_{2}\) solution if the molar conductivity of salt at \(25^{\circ }\text{C}\) is \(325\ \Omega ^{-1}\ \text{cm}^{2}\ \text{mol}^{-1}\).

• A. \(8.125\times 10^{-2}\ \Omega ^{-1}\text{cm}^{-1}\)
• B. \(8.125\times 10^{-3}\ \Omega ^{-1}\text{cm}^{-1}\)
• C. \(7.6923\times 10^{-2}\ \Omega ^{-1}\text{cm}^{-1}\)
• D. \(8.372\times 10^{-2}\ \Omega ^{-1}\text{cm}^{-1}\)

Hint:

Use molar conductivity = 1000 x conductivity / molarity.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
\(\Lambda_m=\dfrac{1000\,\kappa}{M}\) when \(\kappa\) is in \(\Omega^{-1}\text{cm}^{-1}\) and \(M\) is in mol/L.

Step 2: Calculation
\(\kappa=\dfrac{\Lambda_m\times M}{1000}=\dfrac{325\times0.025}{1000}\).

Step 3: Result
\(\kappa=8.125\times10^{-3}\ \Omega^{-1}\text{cm}^{-1}\).

Final Answer:
The conductivity is \(8.125\times10^{-3}\ \Omega^{-1}\text{cm}^{-1}\). \[ \boxed{\text{(B) }8.125\times10^{-3}\ \Omega^{-1}\text{cm}^{-1}} \]