0.6 mL of acetic acid is dissolved in 1 litre of water. The value of van't Hoff factor is 1.04. What will be the degree of dissociation of the acetic acid?

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.04

Explanation

To solve the problem, we need to find the degree of dissociation (α) of acetic acid when it is dissolved in water, given the van't Hoff factor (i) is 1.04.

### Step-by-Step Solution:

1. **Understanding the dissociation of acetic acid**:
Acetic acid (CH₃COOH) dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
This means that for every mole of acetic acid that dissociates, it produces one mole of acetate ions (CH₃COO⁻) and one mole of hydrogen ions (H⁺).

2. **Setting up the initial and final concentrations**:
- Initially, we have 1 mole of acetic acid (since we are considering 0.6 mL of acetic acid, we can convert this to moles if needed, but for the purpose of this calculation, we can assume it as 1 mole for simplicity).
- Let α be the degree of dissociation.
- At equilibrium:
- Moles of CH₃COOH = \(1 - \alpha\)
- Moles of CH₃COO⁻ = α
- Moles of H⁺ = α

3. **Calculating the total number of particles**:
The total number of particles after dissociation will be:
\[ \text{Total particles} = (1 - \alpha) + \alpha + \alpha = 1 + \alpha \]

4. **Using the van't Hoff factor**:
The van't Hoff factor (i) is defined as the ratio of the total number of particles in solution after dissociation to the number of formula units initially dissolved. Therefore:
\[ i = \frac{\text{Total particles}}{\text{Initial particles}} = \frac{1 + \alpha}{1} \]
Given that \(i = 1.04\), we can set up the equation:
\[ 1 + \alpha = 1.04 \]

5. **Solving for α**:
Rearranging the equation gives:
\[ \alpha = 1.04 - 1 = 0.04 \]

### Conclusion:
The degree of dissociation (α) of acetic acid is **0.04**.