Van't Hoff's factor (i) can be calculated using the formula:

i = ΔTF / ΔT0

Where:
ΔTF is the observed depression in freezing point
ΔT0 is the theoretical depression in freezing point for a non-volatile solute

To calculate ΔT0, we can use the formula:

ΔT0 = Kf * m * i

Where:
Kf is the cryoscopic constant for the solvent (in this case, water)
m is the molality of the solute (moles of solute per kilogram of solvent)
i is the van't Hoff's factor

To find m, we need to calculate the number of moles of acetic acid and the mass of water in the solution.

1. Calculating the number of moles of acetic acid:
Given that the volume of acetic acid is 0.6 mL and its density is 1.06 g/cc, we can calculate the mass of acetic acid using the formula:

mass = volume * density
mass = 0.6 mL * 1.06 g/cc
mass = 0.636 g

The molar mass of acetic acid (CH3COOH) is 60.052 g/mol. We can calculate the number of moles using the formula:

moles = mass / molar mass
moles = 0.636 g / 60.052 g/mol
moles = 0.0106 mol

2. Calculating the mass of water:
Given that the volume of water is 1 L, we can calculate the mass of water using its density of 1 g/cc:

mass = volume * density
mass = 1 L * 1 g/cc
mass = 1000 g

3. Calculating the molality of the solute:
molality = moles of solute / mass of solvent (in kg)
molality = 0.0106 mol / 1 kg
molality = 0.0106 mol/kg

Now we can substitute the values into the formula for ΔT0:

ΔT0 = Kf * m * i

Given that the observed depression in freezing point (ΔTF) is 0.0205°C and the cryoscopic constant (Kf) for water is 1.86 Kkg/mol, we can rearrange the formula to solve for i:

i = ΔTF / (Kf * m)

i = 0.0205°C / (1.86 Kkg/mol * 0.0106 mol/kg)
i = 0.0205°C / 0.019686 Kkg/mol
i ≈ 1.04

Therefore, the van't Hoff's factor (i) for this solution is approximately 1.04.