Given,0.6 ml of glacial acetic acid with density 1.06 g/ml is dissolved in 1 kg water and the solution froze at -0.0205 $^{o}C$ . Calculate the van’t Hoff factor:
${{K}_{f}}$ for the water is 1.86 K kg/mol.

Hint: Number of moles of a chemical is the ratio of weight of the chemical to the molecular weight of the chemical.
\[\text{Number of moles of a chemical =}\dfrac{\text{Weight of the chemical}}{\text{molecular weight of the chemical}}\]
The formula to calculate the van’t Hoff factor is as follows.
van’t Hoff factor = $\dfrac{\text{Observed freezing point}}{\text{Calculated freezing point}}$

Complete answer:
- In the question it is given that 0.6 ml of glacial acetic acid is dissolved in 1 kg of water and froze at -0.0205 $^{o}C$ .
- We have to calculate the van’t Hoff factor for the above solution.
- First we have to calculate the weight of the acetic acid and it is as follows.
Weight of the acetic acid = (volume of the acetic acid) (density of the acetic acid)
- Substitute the known values in the above equation to get the weight of the acetic acid.
Volume of the acetic acid = 0.6 ml
Density of the acetic acid = 1.06 g/ml
- Then
 Weight of the acetic acid = (volume of the acetic acid) (density of the acetic acid)
Weight of the acetic acid = (0.6) (1.06) = 0.636 g.
- Number of moles of acetic acid can be calculated as follows.
\[\begin{align}
  & \text{Number of moles of acetic acid =}\dfrac{\text{Weight of the acetic acid}}{\text{molecular weight of the acetic acid}} \\
 & =\dfrac{0.636}{60} \\
 & =0.0106mol \\
\end{align}\]
- From the above data we can calculate the Freezing point of the solution and it is as follows.
\[\Delta {{T}_{f}}=1.86\times 0.0106=0.0197K\]
- Substitute all the known values in the below equation to get the van’t Hoff factor.
\[\begin{align}
  & \text{Van't Hoff factor =}\dfrac{\text{Observed freezing point}}{\text{Calculated freezing point}} \\
 & =\dfrac{0.0205}{0.0197} \\
 & =1.04 \\
\end{align}\]
- Therefore the van’t Hoff factor of the resulting solution is 1.04.

Note:
Acetic acid is a weak acid because it does not dissociate completely into ions when dissolved in water that is why acetic acid is called a weak acid. Strong acid dissociates completely and gives the respective ions in water.