Question

Amount of $75.2g$of $C_6{H}_{5}OH$(phenol) is dissolved in $960g$of solvent of $K_f=14KKgmo{l}^{-1}$. 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

Hint: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 $C_6{H}_{5}OH$
$$\begin{gathered} =6\times 12+5\times 1+1\times 16+1\times 1 \\ =72+5+16+1 \\ =94gmo{l}^{-1} \end{gathered}$$
Molality of the phenolic solution $={\displaystyle \frac{w}{M}}\times {\displaystyle \frac{1000}{massofsolventing}}$
We are provided with these values in the question,
$$\begin{gathered} w=75.2g \\ M=94gmo{l}^{-1} \end{gathered}$$
Mass of solvent $=960g$
Substituting values in the given formula;
$={\displaystyle \frac{75.2}{94}}\times {\displaystyle \frac{1000}{960}}molK{g}^{-1}$
$=0.83molK{g}^{-1}$
Now, we have to calculate the value of Van't Hoff’s factor;
$\mathrm{\Delta }{T}_f=im{K}_f$
We are provided with these values in the question,
$K_f=14KKgmo{l}^{-1}$
$\mathrm{\Delta }{T}_f=7K$
We have calculated molality. It came out to be $=0.83molK{g}^{-1}$
m $=0.83molK{g}^{-1}$
Substituting the values, we get
$7=i\times 14\times 0.83$
$i={\displaystyle \frac{7}{14\times 0.83}}$$=0.602$
The equilibrium reaction for the dimerization of phenol can be written as:
$2C_6{H}_{5}OH\underset{}{\leftrightarrows }{\left(C_6{H}_{5}OH\right)}_2$

(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$$=1-\alpha /2$
Equating it equal to i,
$1-\alpha /2=i$
$$\begin{gathered} 1-i=\alpha /2 \\ 2(1-i)=\alpha \\ 2(1-0.602)=2(0.398)=0.796=0.80=80\mathrm{\%} \end{gathered}$$
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.