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:
Answer & explanation
Correct answer: option 2
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: \( \alpha = 0.5 \).