Practicing Success
Degree of dissociation, when molar conductivity of \(X\) at its concentration \(C\) is \(24.14\) and its limiting molar conductivity is \(48.28\) will be: |
0.3 0.5 0.6 0.9 |
0.5 |
The correct answer is option 2. 0.5. The degree of dissociation (\( \alpha \)) of an electrolyte can be calculated using the formula: \[ \Lambda_m = \Lambda_m^0 \times \alpha \] where: \( \Lambda_m \) is the molar conductivity of the electrolyte at concentration \( C \), \( \Lambda_m^0 \) is the limiting molar conductivity of the electrolyte (at infinite dilution), and \( \alpha \) is the degree of dissociation. Given: \( \Lambda_m = 24.14 \, S \, cm^2 \, mol^{-1} \) \( \Lambda_m^0 = 48.28 \, S \, cm^2 \, mol^{-1} \) We can rearrange the formula to solve for \( \alpha \): \[ \alpha = \frac{\Lambda_m}{\Lambda_m^0} \] \[ \alpha = \frac{24.14 \, S \, cm^2 \, mol^{-1}}{48.28 \, S \, cm^2 \, mol^{-1}} \] \[ \alpha = 0.5 \] So, the correct answer is option 2: \( \alpha = 0.5 \). |