Question

The molar conductivity of \(0.04\) M \(\text{AB}_2\) type salt solution at \(300\) K is \(200\,\Omega ^{-1}\text{cm}^2\text{mol}^{-1}\). Find the conductivity.

• A. \(0.006\,\Omega ^{-1}\text{cm}^{-1}\)
• B. \(0.008\,\Omega ^{-1}\text{cm}^{-1}\)
• C. \(0.01\,\Omega ^{-1}\text{cm}^{-1}\)
• D. \(0.015\,\Omega ^{-1}\text{cm}^{-1}\)

Hint:

kappa = Lambda_m x C / 1000.

Answer 1

The Correct Option is B

Step 1: Units:
$\Lambda_m$ is conductance of the solution containing 1 mol, placed between electrodes 1 cm apart. 1 mol is in $1000/C$ cm$^3$.

Step 2: Compute:
$1000/0.04 = 25000$ cm$^3$ per mole. So $\kappa = \Lambda_m/25000 = 200/25000 = 0.008$.

Step 3: Select:
Option (B).

Final Answer:
Divide molar conductivity by the volume in cm3 per mole. \[ \boxed{B} \]