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
For a weak electrolyte the degree of dissociation α at a concentration c is the ratio of the molar conductivity at that concentration to the limiting molar conductivity:
\( \displaystyle \alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}} = \frac{24.14}{48.28} = 0.5 \)
NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – “… At any concentration c, if α is the degree of dissociation then it can be approximated to the ratio of molar conductivity Λm at the concentration c to limiting molar conductivity, Λm°. Thus we have:
α = Λm / Λm° (2.26)”