Question bank/NEET UG/Chemistry

NEET UGChemistryModerateNEET 2025

If the molar conductivity (Λm\Lambda_m) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be

[Assume Λ+∘=349.6\Lambda_{+}^{\circ} = 349.6 S cm² mol⁻¹ and Λ−∘=50.4\Lambda_{-}^{\circ} = 50.4 S cm² mol⁻¹.]

Show answer and solution
Sol. Degree of dissociation (α\alpha) is given as
α=ΛmΛm∘\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}

Λm∘=Λ+∘+Λ−∘\Lambda_m^\circ = \Lambda_{+}^{\circ} + \Lambda_{-}^{\circ}

= 349.6 + 50.4

= 400 S cm² mol⁻¹

α=ΛmΛm∘=90400=0.225\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{90}{400} = 0.225

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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