??

Asked by SanskarAgarwal86
| 29th Feb, 2020,
04:36: AM
Expert Answer:
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 λm
λm =
Specific conductance, k = Conductance × cell constant
= (1/50)× 0.5 = 0.01 ohm−1 cm−1
λm =
= 200 ohm−1 cm2 mol−1
α =
α =4.21






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