??

Asked by SanskarAgarwal86 | 29th Feb, 2020, 04:36: AM

Expert Answer:

Given:
 
R = 50 Ω
 
Conc. of NH4OH solution = 0.05 M
 
So, conductance = equals 1 over straight R
                         =1 over 50 space ohm to the power of negative 1 end exponent
 
Cell constant = 0.50 cm−1
 
 
Molar conductance λ = 47.4 ohm−1 cm2 mol−1
 
Degree of dissociation,
 
α =straight lambda subscript straight m over straight lambda subscript infinity
 
To calculate λm 
 
λm =fraction numerator straight k cross times 1000 over denominator straight M end fraction
 
Specific conductance, k = Conductance × cell constant
 
                                  = (1/50)× 0.5 = 0.01 ohm−1 cm−1 
 
 
λm =fraction numerator 0.01 cross times 1000 over denominator 0.05 end fraction
     = 200  ohm−1 cm2 mol−1 
 
 
α =fraction numerator 200 over denominator 47.4 end fraction
 
α =4.21 

Answered by Varsha | 2nd Mar, 2020, 12:07: PM