Molar conductivities ( m) at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively. ( m) for CH3COOH will be
- 425.5 S cm2 mol-1
- 180.5 S cm2 mol-1
- 290.8 S cm2 mol-1
- 390.5 S cm2 mol-1
Answer
(4) 390.5 S cm2 mol-1
Solution
By Kohlrausch's law of independent migration of ions:
m(CH3COOH) = m(CH3COONa) + m(HCl) - m(NaCl)
= 91.0 + 425.9 - 126.4
= 390.5 S cm2 mol-1
Answer: (4)