Read the passage carefully and answer.
The molar conductivity of a solution at any given concentration is the conductance of the volume of solution containing one mole of electrolyte kept between two platinum electrodes with unit area of cross-section and at a distance of unit length. Both conductivity and molar conductivity change with the concentration of the electrolyte. Kohlrausch examined \( Ʌ°_m \) values for a number of strong electrolytes and observed certain regularities. He noted that the difference in \( Ʌ°_m \) of the electrolytes NaX and KX for any X is nearly constant. On the basis of the above observations, he enunciated the Kohlrausch law of independent migration of ions.
A \( 0.001\, mol\, L^{-1} \) solution of acetic acid shows conductivity of \( 5.02 × 10^{-5}\, S\, cm^{-1} \). If \( Ʌ°_m \) for acetic acid is \( 390\, cm^2\, mol^{-1} \), then the degree of dissociation is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.12
For a weak electrolyte the degree of dissociation is the ratio of the molar conductivity at the given concentration to the limiting molar conductivity:
\( \displaystyle \alpha = \frac{\Lambda_m}{\Lambda^{\circ}_m} \)
First find \( \Lambda_m \) from the conductivity. With \( \kappa \) in S cm−1 and c in mol L−1, the factor 1000 converts litres to cm3:
\( \displaystyle \Lambda_m = \frac{\kappa \times 1000}{c} \)
\( \displaystyle \Lambda_m = \frac{5.02 \times 10^{-5} \times 1000}{0.001} = 50.2\ \mathrm{S\,cm^{2}\,mol^{-1}} \)
Then:
\( \displaystyle \alpha = \frac{50.2}{390} \approx 0.129 \)
Of the given values, this is closest to 0.12.
Note 1: It seems that there is a mistake in the original NTA paper, which prints the unit of the limiting molar conductivity as cm2 mol−1 when it should read S cm2 mol−1.
Note 2: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Variation of Conductivity and Molar Conductivity with Concentration – “… The conductivity of 0.001028 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate its dissociation constant if Λ°m for acetic acid is 390.5 S cm2 mol–1.”