for NaBr, HBr and CH3COONa are 128.2, 427.7 and 91.0 S cm2 mol–1 respectively. The for CH­3COOH will be464.9 S cm2 mol–1646.9 S cm2 mol–1208.5 S cm2 mol–1390.5 S cm2 mol–1

It seems like you're asking for the molar conductivity (Λm) of CH3COOH (acetic acid) given the molar conductivities of NaBr, HBr, and CH3COONa (sodium acetate).

The molar conductivity of a weak electrolyte at infinite dilution can be calculated using the Kohlrausch's Law of Independent Migration of Ions. According to this law, the molar conductivity of a weak electrolyte at infinite dilution is the sum of the molar conductivities of its ions.

For acetic acid (CH3COOH), it ionizes into CH3COO- and H+ ions.

The molar conductivity of CH3COOH (Λm°(CH3COOH)) can be calculated as follows:

Λm°(CH3COOH) = Λm°(CH3COO-) + Λm°(H+)

We know that CH3COO- comes from CH3COONa and H+ comes from HBr. So, we can write:

Λm°(CH3COOH) = Λm°(CH3COONa) + Λm°(HBr) - Λm°(NaBr)

Substituting the given values:

Λm°(CH3COOH) = 91.0 + 427.7 - 128.2 = 390.5 S cm2 mol–1

So, the molar conductivity of CH3COOH at infinite dilution is 390.5 S cm2 mol–1.