??

Given:

 

R = 50 Ω

 

Conc. of NH₄OH solution = 0.05 M

 

So, conductance = 

                         =

 

Cell constant = 0.50 cm⁻¹

 

 

Molar conductance λ_∞ = 47.4 ohm⁻¹ cm² mol⁻¹

 

Degree of dissociation,

 

α =

 

To calculate λ_m 

 

λ_m =

 

Specific conductance, k = Conductance × cell constant

 

                                  = (1/50)× 0.5 = 0.01 ohm⁻¹ cm⁻¹ 

 

 

λ_m =

     = 200  ohm⁻¹ cm² mol⁻¹ 

 

 

α =

 

α =4.21