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Question: Given,0.6 ml of glacial acetic acid with density 1.06 g/ml is dissolved in 1 kg water and the soluti...

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 oC^{o}C . Calculate the van’t Hoff factor:
Kf{{K}_{f}} for the water is 1.86 K kg/mol.

Explanation

Solution

Number of moles of a chemical is the ratio of weight of the chemical to the molecular weight of the chemical.
Number of moles of a chemical =Weight of the chemicalmolecular 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 = Observed freezing pointCalculated freezing point\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 oC^{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.

& \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{aligned}$$ \- From the above data we can calculate the Freezing point of the solution and it is as follows. $$\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{aligned} & \text{Van't Hoff factor =}\dfrac{\text{Observed freezing point}}{\text{Calculated freezing point}} \\\ & =\dfrac{0.0205}{0.0197} \\\ & =1.04 \\\ \end{aligned}$$ \- 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.