Question

Calculate limiting molar conductivity of aloh 3

Answer 1

Explanation:

according to Kohlrausch's law, molar conductivity at infinite dilution can be broken into its ions.

so, for Al(OH)3 = for Al^+ + 2 × for (OH)^-

= 1/2 × for Al2(SO4)3 + 3 × for NH4(OH) - 3/2 × for (NH4)2SO4.

for Al2(SO4)3 = 858 S.cm²/mol.

for NH4OH = 238.3 S cm²/mol.

for (NH4)2SO4 = 238.4 S cm²/mol.