Question

Amount of $75.2\,g$of ${C_6}{H_5}OH$(phenol) is dissolved in $960\,g$of solvent of ${K_f} = 14K\,Kg\,mo{l^{ - 1}}$. If the depression in the freezing point is 7K, then find the percentage of phenol that dimerizes.
A) $75$
B) $80$
C) $30$
D) $40$

Answer 1

Freezing point is defined as the temperature at which the chemical potential of a pure liquid solvent is equal to the chemical potential of a solid solvent, upon addition of a non-volatile solute into the solvent the chemical potential value falls and now the liquid solvent reaches the same chemical potential value as that of solid solvent at a temperature lower than the earlier this in scientific terms is known as depression in freezing point. Depression in Freezing point is a colligative property thus it only depends on the number of non- volatile solute particles added and not on their type.

Complete solution:
Calculating the molar mass of the solute,
Molar mass of phenol ${C_6}{H_5}OH$
$= 6 \times 12 + 5 \times 1 + 1 \times 16 + 1 \times 1 \\\ = 72 + 5 + 16 + 1 \\\ = 94\,g\,mo{l^{ - 1}} \\\$
Molality of the phenolic solution $= \dfrac{w}{M} \times \dfrac{{1000}}{{mass\,of\,solvent\,in\,g}}$
We are provided with these values in the question,
$w = 75.2\,g \\\ M = 94\,g\,mo{l^{ - 1}} \\\ \\\$
Mass of solvent $= 960g$
Substituting values in the given formula;
$= \dfrac{{75.2}}{{94}} \times \dfrac{{1000}}{{960}}mol\,K{g^{ - 1}}$
$= 0.83\,mol\,K{g^{ - 1}}$
Now, we have to calculate the value of Van't Hoff’s factor;
$\Delta {T_f} = im{K_f}$
We are provided with these values in the question,
${K_f} = 14K\,Kg\,mo{l^{ - 1}}$
$\Delta {T_f} = 7K$
We have calculated molality. It came out to be $= 0.83\,mol\,K{g^{ - 1}}$
m $= 0.83\,mol\,K{g^{ - 1}}$
Substituting the values, we get
$7 = i \times 14 \times 0.83$
$i = \dfrac{7}{{14 \times 0.83}}$ $= 0.602$
The equilibrium reaction for the dimerization of phenol can be written as:
$2{C_6}{H_5}OH\underset {} \leftrightarrows {\left( {{C_6}{H_5}OH} \right)_2}$

(initially when whole phenol is non-dimerized)	$1$	$0$
(degree of association $\alpha$)	$1 - \alpha$	$\alpha /2$

Total no. of moles $= 1 - \alpha + \alpha /2$ $= 1 - \alpha /2$
Equating it equal to i,
$1 - \alpha /2 = i$
$1 - i = \alpha /2 \\\ 2(1 - i) = \alpha \\\ 2(1 - 0.602) = 2(0.398) = 0.796 = 0.80 = 80\% \\\$
So, the correct answer is $\left( b \right)80$.

Note: While doing questions like above where dimerization is involved, always use the degree of association wisely because even a small error while handling the degree of dissociation would change the answer completely leading to a different incorrect answer. Also be careful while calculating molar mass of the solute.