NCERT Exercise 2.11 - Electrochemistry
Question: Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?
Answer: **NCERT answer:** 1.85 × 10–5
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