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.
Answer
607.8k+ views
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.
\[\text{Number of moles of a chemical =}\dfrac{\text{Weight of the chemical}}{\text{molecular weight of the chemical}}\]
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.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Name the crygenes that control cotton bollworm and class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many molecules of ATP and NADPH are required information class 12 biology CBSE

In a transcription unit the promoter is said to be class 12 biology CBSE

