KEAM 2011 Physics and Chemistry Paper

According to Kohlrausch's law,
∧∘ for CH3COOH=λCH3COO−∘+λH+∘
∧∘ for NaCl=λNa+∘+λCl−∘
=125‌mho‌cm2‌mol−1....(i)
∧∘ for HCl=λH+∘+λCl−∘
=425‌mho‌cm2‌mol−1...(2)
∧∘ for CH3COONa=λCH3COO−∘+λNa+∘
=90‌mho‌cm2‌mol−1
Adding Eqs. (ii) and (iii) and subtracting (i), we get
=425+90−125
=390‌mho‌cm2‌mol−1
or∧∘CH3COOH=λCH3COO−+λH+∘
=390‌mho‌cm2‌mol−1
Thus, the molar conductivity of CH3COOH at infinite dilution ∧∘=390‌mho‌cm2‌mol−1
The molar conductivity of 0.1MCH3COOH solution (∧mc)
=7.8‌mho‌cm2‌mol−1
Degree of dissociation (α)=‌

∧mc

∧m∘

=‌

7.8

390

=0.02