Question

75.2g of ${{\text{C}}_{6}}{{\text{H}}_{5}}\text{OH}$ (phenol) is dissolved in a solvent in ${{\text{K}}_{\text{f}}}\ \text{= 14}$. If the depression in freezing point is 7K then find the % of phenol that of dimerise.

Answer 1

For this problem, we have to write the reaction of the dimerization of the phenol then by determining the moles of the reactant and product molecules, we can calculate the molality. Finally, we will apply the formula of depression in freezing point.

Complete Step-by-Step answer:
- In the given question, we have to calculate the percentage of phenol that will dimerise.
- As we know that the dimerization is the process in which two molecules either same or different combine to form the dimer.
- Now, the reaction of the dimerization of the phenol is:
$\text{2}{{\text{C}}_{6}}{{\text{H}}_{6}}\text{OH }\to \text{ (}{{\text{C}}_{6}}{{\text{H}}_{6}}\text{OH}{{\text{)}}_{2}}$
- Now, initially, let the number of moles of phenol is C then the number of moles of the product will be 0.
- After some time, the reaction will complete and the number of moles of the reactant will become $\text{C(1 - }\alpha \text{)}$ and the number of moles of product will be C$\alpha$/2.
- So, the total number of moles will be: $\text{C - C}\alpha \text{ + C}\alpha /2\text{ = }\dfrac{\text{C(2 - }\alpha \text{)}}{2}$ …..(1)
- Now, as we know that the molecular weight of the phenol will be:
$12\text{ }\times \text{ 6 + 5 }\times \text{ 1 + 16 + 1 = 94}$
- So, the number of moles of the phenol (C) will be the ratio of mass to the molecular weight i.e. $\dfrac{\text{75}\text{.2}}{94}\ \text{= 0}\text{.8}$
- So, equation (1) becomes, $\dfrac{\text{0}\text{.8(2 - }\alpha \text{)}}{2}$
- Now, we will apply the depression in freezing point we will get the value of alpha or the percentage of phenol i.e.
$\vartriangle {{\text{T}}_{\text{f }}}=\text{ }{{\text{K}}_{f}}\text{ }\text{+ m}$
- Here, the value of depression in freezing point is 7 K and ${{\text{K}}_{\text{f}}}\ \text{= 14}$
$7\ \text{ = 14 }\text{+ 0}\text{.8}\left( \dfrac{2-\alpha }{2} \right)$
$1\ \text{ = 0}\text{.8}\left( 2-\alpha \right)$
$1.25\ \text{ = }\left( 2-\alpha \right)$
$\alpha \ =\text{ }\text{0}\text{.75}$ or 75%.
Therefore, 75% of phenol will dimerise.

Note: In the formula of depression of freezing point, $\vartriangle {{\text{T}}_{\text{f}}}$ is the depression in freezing point, ${{\text{K}}_{f}}$ is the freezing point depression constant and m is the molality. Whereas molality is defined as the ratio of number of moles of solute to the mass of the solvent.