NEETChemistryElectrochemistry
The molar conductivity of 0.00241 M acetic acid is 7.896 × 10 -5 S cm -1 . What is its molar conductivity? If 0 Λ m for acetic acid is 390.5 S cm .2 mol -1 , what is its dissociation constant?
Options
- AMolar conductivity = 33 S cm 2 mol -1 , Dissociation constant = 1.86 × 10 -5
- BMolar conductivity = 7.896 × 10 -5 S cm 2 mol -1 , Dissociation constant = 390.5
- CMolar conductivity = 390.5 S cm 2 mol -1 , Dissociation constant = 7.896 × 10 -5
- DMolar conductivity = 1.86 × 10 -5 S cm 2 mol -1 , Dissociation constant = 33
Correct answer
B. Molar conductivity = 7.896 × 10 -5 S cm 2 mol -1 , Dissociation constant = 390.5
Step-by-step solution
Correct Option is : (B) Molar conductivity = 7.896 × 10 -5 S cm 2 mol -1 , Dissociation constant = 390.5