NTA Abhyas NEET2020ChemistryElectrochemistryPractice
The specific conductance of a 0.5 N solution of an electrolyte at 25 ° C is 0.00045 S c m - 1 . The equivalent conductance of this electrolyte at infinite dilution is 300 S c m 2 e q - 1 . The degree of dissociation of the electrolyte is
Options
- A0.66
- B0.03
- C0.003
- D0.3
Correct answer
C. 0.003
Step-by-step solution
Λ v = K × 1000 C N = 0.00045 × 1000 0.5 = 0.9 S c m 2 e q - 1 α = Λ v Λ ∞ = 0.9 300 = 3 1000 = 0.003