
Calculate van’t Hoff factor i of the compound acetic acid in benzene if their alpha=50%.
a.) 0.25
b.) 1.5
c.) 0.75
d.) 2
Answer
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Hint: In order to deal with this question first we will define the term van’t Hoff factor , further then we will calculate the van’t Hoff factor of the given compound by using the formula which is mentioned in solution.
Formula used- $i = 1 - \alpha + \dfrac{\alpha }{n}$
Complete step-by-step answer:
Van't Hoff factor: The van't Hoff factor I is a measure of the influence of a liquid on colligative properties such as osmotic pressure, relative vapor pressure reduction, elevation of the boiling point and deflation of the freezing point.
Given that $\alpha = 50\% = \dfrac{1}{2}$
Acetic acid and benzene reaction is not possible here. But yes, in the case of benzene, acetic acid dimers, interaction occurs.
Here two mole of acetic acid produce 1 mole of dimer acetic acid .
So, $n = 2$
We know the formula of van’t Hoff factor which is given as
van't Hoff's factor
$i = 1 - \alpha + \dfrac{\alpha }{n}$
$\alpha = 50\% = \dfrac{1}{2}$ (given )
Substitute the value of $\alpha = 50\% = \dfrac{1}{2}$ and n in above formula we have
$
\because i = 1 - \alpha + \dfrac{\alpha }{n} \\
\Rightarrow i = 1 - \dfrac{1}{2} + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{2} \\
\Rightarrow i = 1 - \dfrac{1}{2} + \dfrac{1}{4} \\
\Rightarrow i = 1 - 0.5 + 0.25 = 0.75 \\
$
Hence, the van’t Hoff factor is equal to 0.75
So, the correct answer is option C.
Note- The van't Hoff effect is the ratio between the real concentration of particles generated when dissolving the material and the concentration of a substance determined from its density. The van't Hoff factor for most non-electrolytes dissolved in water is approximately 1. For most ionic compounds dissolved in water the element van't Hoff is proportional to the number of distinct ions in the substance's formula unit. This is only valid for optimal solutions, as pairing of ions happens periodically in solution. A small percentage of the ions are combined at a given moment, and count as a single electron.
Formula used- $i = 1 - \alpha + \dfrac{\alpha }{n}$
Complete step-by-step answer:
Van't Hoff factor: The van't Hoff factor I is a measure of the influence of a liquid on colligative properties such as osmotic pressure, relative vapor pressure reduction, elevation of the boiling point and deflation of the freezing point.
Given that $\alpha = 50\% = \dfrac{1}{2}$
Acetic acid and benzene reaction is not possible here. But yes, in the case of benzene, acetic acid dimers, interaction occurs.
Here two mole of acetic acid produce 1 mole of dimer acetic acid .
So, $n = 2$
We know the formula of van’t Hoff factor which is given as
van't Hoff's factor
$i = 1 - \alpha + \dfrac{\alpha }{n}$
$\alpha = 50\% = \dfrac{1}{2}$ (given )
Substitute the value of $\alpha = 50\% = \dfrac{1}{2}$ and n in above formula we have
$
\because i = 1 - \alpha + \dfrac{\alpha }{n} \\
\Rightarrow i = 1 - \dfrac{1}{2} + \dfrac{{\left( {\dfrac{1}{2}} \right)}}{2} \\
\Rightarrow i = 1 - \dfrac{1}{2} + \dfrac{1}{4} \\
\Rightarrow i = 1 - 0.5 + 0.25 = 0.75 \\
$
Hence, the van’t Hoff factor is equal to 0.75
So, the correct answer is option C.
Note- The van't Hoff effect is the ratio between the real concentration of particles generated when dissolving the material and the concentration of a substance determined from its density. The van't Hoff factor for most non-electrolytes dissolved in water is approximately 1. For most ionic compounds dissolved in water the element van't Hoff is proportional to the number of distinct ions in the substance's formula unit. This is only valid for optimal solutions, as pairing of ions happens periodically in solution. A small percentage of the ions are combined at a given moment, and count as a single electron.
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