Question

The specific conductance of 0.01 M aqueous solution of acetic acid is 1. 63 x10-2 S m^-1 and the molar conductance at infinite dilution is 390.7x10^-4 S m² mol^-1.Calculate the degree of dissociation of the acetic acid solution?

• A. 0.4
• B. 0.239
• C. 0.0417
• D. 0.0239

Answer 1

The Correct Option is C

To calculate the degree of dissociation (α) of the acetic acid solution, we'll follow these steps:

1. Given Data:
- Specific conductance (κ) = 1.63 × 10^-2 S m^-1
- Molar concentration (c) = 0.01 M = 0.01 × 10³ mol m^-3 = 10 mol m^-3
- Molar conductance at infinite dilution (Λ⁰) = 390.7 × 10^-4 S m² mol^-1

2. Calculate Molar Conductance (Λ_m):
Molar conductance is given by:
Λ_m = κ / c
= (1.63 × 10^-2 S m^-1) / (10 mol m^-3)
= 1.63 × 10^-3 S m² mol^-1

3. Calculate Degree of Dissociation (α):
The degree of dissociation is given by:
α = Λ_m / Λ⁰
= (1.63 × 10^-3 S m² mol^-1) / (390.7 × 10^-4 S m² mol^-1)
= (1.63 × 10^-3) / (3.907 × 10^-2)
= 0.0417 or 4.17%

4. Interpretation:
The low degree of dissociation (4.17%) indicates that acetic acid is a weak electrolyte, as expected, with only a small fraction of molecules dissociated into ions at this concentration.

Final Answer:
The degree of dissociation of the 0.01 M acetic acid solution is 0.0417 (or 4.17%).