
Acetic acid dimerises in Benzene solution. The Van’t Hoff factor for the dimerization of Acetic acid is \[0.8\]. The percentage of dimerization of Acetic acid will be:
A. \[20\% \]
B. \[40\% \]
C. \[60\% \]
D. \[80\% \]
Answer
490.8k+ views
Hint: When Acetic acid is added to the solution of Benzene the dimerization of acid takes place. Also the Van’t Hoff factor is given for this dimerization of acid. Thus by using the relation between the van't Hoff factor and degree of association we can find the percentage of dimerization for acetic acid.
Formula used:
\[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
Where, \[i\] is known as the Van't Hoff factor, \[\alpha \] is known as the degree of association and \[n\] is the atomicity of a product. Here \[n\] is equal to two.
Complete answer:
When acetic acid is added with benzene then a dimer of acetic acid is formed. This can be represented by using the chemical representation as,
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH }} \rightleftharpoons {\text{ }}{\left( {{\text{ C}}{{\text{H}}_3}{\text{COOH}}} \right)_2}\]
For the above reaction, we can use a relation between the degree of association and the Van't Hoff factor.
\[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
Here, we are given with \[i{\text{ = 0}}{\text{.8}}\] and since it forms a dimer so the value of \[n\] is two. \[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
On substituting the values we get the result as,
\[{\text{0}}{\text{.8 = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{2}}}{1}\]
\[\Rightarrow \alpha {\text{ = 0}}{\text{.4}}\]
Hence the degree of association is \[{\text{0}}{\text{.4}}\] , for finding the percentage of degree of association we can convert it into percentage as,
Percentage of association\[{\text{ = 0}}{\text{.4 }} \times {\text{ 100}}\]
Percentage of association \[{t{ = 40\% }}\]
Therefore, the percentage of association is \[{{ 40\% }}\]. Therefore, the correct option is option (B).
Note:
\[\alpha \] can be a degree of association or degree of dissociation. It depends on the reaction whether association takes place or dissociation takes place. The relation which we use in the above problem is valid for such reactions only. We can also derive the relation by calculating the total number of moles after some extent of association. When the particle of solute gets associated, \[i\] will be less than one. When solute particles dissociate then the value of \[i\] will be greater than one.
Formula used:
\[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
Where, \[i\] is known as the Van't Hoff factor, \[\alpha \] is known as the degree of association and \[n\] is the atomicity of a product. Here \[n\] is equal to two.
Complete answer:
When acetic acid is added with benzene then a dimer of acetic acid is formed. This can be represented by using the chemical representation as,
\[{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH }} \rightleftharpoons {\text{ }}{\left( {{\text{ C}}{{\text{H}}_3}{\text{COOH}}} \right)_2}\]
For the above reaction, we can use a relation between the degree of association and the Van't Hoff factor.
\[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
Here, we are given with \[i{\text{ = 0}}{\text{.8}}\] and since it forms a dimer so the value of \[n\] is two. \[i{\text{ = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{n}}}{1}\]
On substituting the values we get the result as,
\[{\text{0}}{\text{.8 = }}\dfrac{{1 - \alpha {\text{ + }}\dfrac{\alpha }{2}}}{1}\]
\[\Rightarrow \alpha {\text{ = 0}}{\text{.4}}\]
Hence the degree of association is \[{\text{0}}{\text{.4}}\] , for finding the percentage of degree of association we can convert it into percentage as,
Percentage of association\[{\text{ = 0}}{\text{.4 }} \times {\text{ 100}}\]
Percentage of association \[{t{ = 40\% }}\]
Therefore, the percentage of association is \[{{ 40\% }}\]. Therefore, the correct option is option (B).
Note:
\[\alpha \] can be a degree of association or degree of dissociation. It depends on the reaction whether association takes place or dissociation takes place. The relation which we use in the above problem is valid for such reactions only. We can also derive the relation by calculating the total number of moles after some extent of association. When the particle of solute gets associated, \[i\] will be less than one. When solute particles dissociate then the value of \[i\] will be greater than one.
Recently Updated Pages
The number of solutions in x in 02pi for which sqrt class 12 maths CBSE

Write any two methods of preparation of phenol Give class 12 chemistry CBSE

Differentiate between action potential and resting class 12 biology CBSE

Two plane mirrors arranged at right angles to each class 12 physics CBSE

Which of the following molecules is are chiral A I class 12 chemistry CBSE

Name different types of neurons and give one function class 12 biology CBSE

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

What are the major means of transport Explain each class 12 social science CBSE

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

Differentiate between insitu conservation and exsitu class 12 biology CBSE

State the principle of an ac generator and explain class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

