Phenol associate in benzene as
\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons \dfrac{1}{2}{({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}})_2}\]
If the degree of association of phenol is \[{{40\% }}\]. Van’t Hoff factor is:
A) 1
B) 0.8
C) 1.4
D) 0.6
Answer
599.1k+ views
Hint:Degree of association is defined as the fraction of the total number of molecules of the substance which exists as associated molecules. Using the given value of percent degree of association calculates the value of the degree of association of phenol. Finally using the equation related to the degree of association and Van’t Hoff factor calculate the value of Van’t Hoff factor.
Formula Used: \[i = 1 - \alpha \left( {1 - \dfrac{1}{n}} \right)\]
Complete answer:
We have given the percent degree of association of phenol is \[{\text{40\% }}\]. So, the degree of association, \[\alpha = \dfrac{{40}}{{100}} = 0.4\]
Phenol associate in benzene as
\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons \dfrac{1}{2}{({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}})_2}\]
From the given reaction we can say that phenol is associated as a dimer. So, the factor of association, \[n = 2\]
Now, we can calculate the Van’t Hoff factor as follows:
\[i = 1 - \alpha \left( {1 - \dfrac{1}{n}} \right)\]
Where,
\[i\]= Van’t Hoff factor
Now, substitute 0.4 for \[\alpha \] and 2 for \[n\] and calculate the Van’t Hoff factor.
\[i = 1 - 0.4\left( {1 - \dfrac{1}{2}} \right)\]
\[i = 0.8\]
Hence, the Van't Hoff factor is 0.8. I.e, option ‘B’ is the correct answer
Note:There are two different equations used to calculate the Van’t Hoff factor depending on the degree of association and degree of dissociation. So we have to use the correct equation to calculate the Van’t Hoff factor. Degree of association refers to the fraction of the total number of molecules which associate or combine together resulting in the formation of bigger molecules. While the degree of dissociation refers to the fraction of total molecules that dissociate into simpler molecules.
Formula Used: \[i = 1 - \alpha \left( {1 - \dfrac{1}{n}} \right)\]
Complete answer:
We have given the percent degree of association of phenol is \[{\text{40\% }}\]. So, the degree of association, \[\alpha = \dfrac{{40}}{{100}} = 0.4\]
Phenol associate in benzene as
\[{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}} \rightleftharpoons \dfrac{1}{2}{({{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}})_2}\]
From the given reaction we can say that phenol is associated as a dimer. So, the factor of association, \[n = 2\]
Now, we can calculate the Van’t Hoff factor as follows:
\[i = 1 - \alpha \left( {1 - \dfrac{1}{n}} \right)\]
Where,
\[i\]= Van’t Hoff factor
Now, substitute 0.4 for \[\alpha \] and 2 for \[n\] and calculate the Van’t Hoff factor.
\[i = 1 - 0.4\left( {1 - \dfrac{1}{2}} \right)\]
\[i = 0.8\]
Hence, the Van't Hoff factor is 0.8. I.e, option ‘B’ is the correct answer
Note:There are two different equations used to calculate the Van’t Hoff factor depending on the degree of association and degree of dissociation. So we have to use the correct equation to calculate the Van’t Hoff factor. Degree of association refers to the fraction of the total number of molecules which associate or combine together resulting in the formation of bigger molecules. While the degree of dissociation refers to the fraction of total molecules that dissociate into simpler molecules.
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