Question
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 . Calculate the van’t Hoff factor:
Kf for the water is 1.86 K kg/mol.
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 =molecular weight of the chemicalWeight of the chemical
The formula to calculate the van’t Hoff factor is as follows.
van’t Hoff factor = Calculated freezing pointObserved 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 .
- 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.