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Hint: Try to recall about the concept of Van’t Hoff factor which explains the abnormality in colligative properties when a solute present in solution associates or dissociates. Now, by using this you can easily solve the question given above.
Complete answer:
It is known to you that the Van't Hoff factor is used in the determination of certain colligative properties (properties which depend on the relative number of solvent and solute molecules).
Van’t Hoff factor is used when electrolytic solute is present in solution because the number of particles would be different then the number of particles actually added, due to dissociation or association of solutes.
If solute gets associated or dissociated in solution then experimental/observed/actual value of colligative property will be different from theoretically predicted value so it is also known as abnormal colligative property.
This abnormality can be calculated in terms of Van’t Hoff factor:
$i = \dfrac{{actual{\text{ number of particles or concentration}}}}{{theoretical{\text{ number of particles or concentration}}}}$.
Now, calculating Van’t Hoff factor for ethanoic acid in benzene:
$2C{H_3}COOH \to {\left( {C{H_3}COOH} \right)_2}$
It is known to you that ethanoic acid dimerizes 100% in benzene. So, $\alpha $=degree of association of ethanoic acid in benzene=1
$
C{\text{ 0}}
{\text{C(1 - }}\alpha {\text{) }}\dfrac{{{\text{C}}\alpha }}{2}
\Rightarrow Net{\text{ concentration = }}C - C\alpha + \dfrac{{{\text{C}}\alpha }}{2}
\Rightarrow i = \dfrac{{C - C\alpha + \dfrac{{{\text{C}}\alpha }}{2}}}{C}
\Rightarrow i = 1 + \left( {\dfrac{1}{2} - 1} \right)\alpha
\therefore i = \dfrac{1}{2}\left[ {\because \alpha = 1} \right]
$
Hence, from above calculation we can say that the value of Van’t Hoff factor for ethanoic acid in benzene is i=1/2.
Note: It should be remembered to you that colligative properties are those properties of solution which are dependent only on the total number of particles or total concentration of particles in the solution and are not dependent on the nature of particle i.e. shape, size etc. of the particles.
Complete answer:
It is known to you that the Van't Hoff factor is used in the determination of certain colligative properties (properties which depend on the relative number of solvent and solute molecules).
Van’t Hoff factor is used when electrolytic solute is present in solution because the number of particles would be different then the number of particles actually added, due to dissociation or association of solutes.
If solute gets associated or dissociated in solution then experimental/observed/actual value of colligative property will be different from theoretically predicted value so it is also known as abnormal colligative property.
This abnormality can be calculated in terms of Van’t Hoff factor:
$i = \dfrac{{actual{\text{ number of particles or concentration}}}}{{theoretical{\text{ number of particles or concentration}}}}$.
Now, calculating Van’t Hoff factor for ethanoic acid in benzene:
$2C{H_3}COOH \to {\left( {C{H_3}COOH} \right)_2}$
It is known to you that ethanoic acid dimerizes 100% in benzene. So, $\alpha $=degree of association of ethanoic acid in benzene=1
$
C{\text{ 0}}
{\text{C(1 - }}\alpha {\text{) }}\dfrac{{{\text{C}}\alpha }}{2}
\Rightarrow Net{\text{ concentration = }}C - C\alpha + \dfrac{{{\text{C}}\alpha }}{2}
\Rightarrow i = \dfrac{{C - C\alpha + \dfrac{{{\text{C}}\alpha }}{2}}}{C}
\Rightarrow i = 1 + \left( {\dfrac{1}{2} - 1} \right)\alpha
\therefore i = \dfrac{1}{2}\left[ {\because \alpha = 1} \right]
$
Hence, from above calculation we can say that the value of Van’t Hoff factor for ethanoic acid in benzene is i=1/2.
Note: It should be remembered to you that colligative properties are those properties of solution which are dependent only on the total number of particles or total concentration of particles in the solution and are not dependent on the nature of particle i.e. shape, size etc. of the particles.
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