Question

The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g of benzene is lowered by ${{0}}{{.4}}{{{5}}^{{0}}}{{C}}$. Calculate the degree of association of acetic acid in benzene:
(${{{K}}_{{f}}}$ for benzene $ = $ 5.12 kg/mol)
A.5.5 $\% $
B.64.5 $\% $
C.35.5 $\% $
D.94.6 $\% $

Answer 1

Hint:Colligative properties in chemistry are defined as those properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution and not on the nature of the chemical species that are present in the solution. Depression in freezing is one of the colligative properties.

Complete step by step answer:
On the addition of a solute the freezing point of the solvent decreases and this property is known as depression in freezing point. And it is directly proportional to the molality of the solute.
$\Rightarrow$${{\Delta }}{{{T}}_{{f}}}{{ = T}}_{{f}}^{{0}}{{ - }}{{{T}}_{{f}}}{{ = i}}{{{K}}_{{f}}}{{m}}$
$\Rightarrow$${{\Delta }}{{{T}}_{{f}}}$ is the depression in the freezing point
$\Rightarrow$\[{{T}}_{{f}}^{{0}}\] is the freezing point of the pure solvent
$\Rightarrow$\[{{{T}}_{{f}}}\] is the freezing point of the solution
I is the van’t Hoff’s factor
m is molality and \[{{{K}}_{{f}}}\] is depression constant.
Now, molality of the solution is $ = $ \[\dfrac{{\dfrac{{{{0}}{{.2}}}}{{{{60}}}}}}{{\dfrac{{{{20}}}}{{{{1000}}}}}}\] $ = $ 0.166 m
We are provided with ${{\Delta }}{{{T}}_{{f}}} = {{0}}{{.4}}{{{5}}^{{0}}}{{C}}$ and \[{{{K}}_{{f}}}\] is 5.12
Now, putting all the values in the equation we have,
$\Rightarrow$\[{{0}}{{.45 = i \times 5}}{{.12 \times 0}}{{.166}}\], which gives us i $ = $ 0.527.
Acetic acid undergoes dimerization in benzene:
$\Rightarrow$\[{{2C}}{{{H}}_{{3}}}{{COOH}} \rightleftharpoons {{{(C}}{{{H}}_{{3}}}{{COOH)}}_{{2}}}\]
If x is the degree of association of solute, then we would have ($1 - x$) mol of acetic acid left in unassociated form and x/s as associated form.
Therefore, the total number of moles of particles at equilibrium:
$\Rightarrow$\[{{i = 1 - x + }}\dfrac{{{x}}}{{{2}}}\], which gives x $ = $ 0.946.
Which gives the percentage of association to be 94.6 $\% $

Therefore, the correct answer is D.

Note:
The van’t Hoff factor is the relation between the actual concentration of particles produced when the substance is dissolved and the concentration of the substance as calculated from its mass. The van’t Hoff factor for a non-electrolytic solute is 1. It will be greater than 1 if the solute undergoes dissociation in the solvent and lesser than 1 when the solute undergoes association in the solvent.