Question

The molecular mass of benzoic acid as benzene as determined by depression in freezing point method corresponds to:
A. ionization of benzoic acid
B. dimerization of benzoic acid
C. trimerization of benzoic acid
D. solvation of benzoic acid

Answer 1

Hint:The molecular mass of any compound can be determined by the sum total of the atomic weight of all the atoms present in the compound. The actual molecular mass of benzoic acid is 122 g/mol. The observed molecular weight of benzoic acid by the method of depression in freezing point is 244 g/mol.

Complete step by step answer:The chemical formula of benzoic acid is ${C_6}{H_5}COOH$.
To calculate the molecular weight of benzoic acid multiply the number of atoms with the atomic weight of the atom and add all the values obtained.
The calculation for the molecular weight of benzoic acid is shown below.
$= (6 \times 12) + (5 \times 1) + 12 + 6 + 16 + 1$
$= 122\;{\text{g}}/{\text{mol}}$
The depression in freezing point is a colligative property which determines the lowering in the freezing point value of the solvent by the addition of solute.
The depression in the freezing point is calculated by the formula as shown below.
$\Delta {T_f} = \dfrac{{1000 \times {K_f} \times {w_2}}}{{M \times {w_1}}}$
Where,
$\Delta {T_f}$ is the depression in freezing point.
${K_f}$ is the molal depression constant.
${w_1}$ is the weight of the solvent.
${w_2}$ is the weight of solute.
M is the molecular weight.
Thus, the molecular weight of the compound can be calculated by this formula.
The observed molecular weight of benzoic acid is 244 g/mol.
The Van't Hoff factor (i) is calculated as the formula shown below.
$i = \dfrac{{Actual\;value}}{{Observed\;value}}......(i)$
Substitute the values of the actual molecular weight of benzoic acid and observed weight of benzoic acid in the above equation.
$\Rightarrow i = \dfrac{{122}}{{244}}$
$\Rightarrow i = \dfrac{1}{2}$
The benzoic acid changes from
$i = \dfrac{1}{n} = \dfrac{1}{2}$
$n{C_6}{H_5}COOH \to {({C_6}{H_5}COOH)_n}.......(ii)$
$initial\;\;\;1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;0$
$final\;\;\;\;0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\dfrac{1}{n}$
$n = 2$
Substitute the value in equation (ii), we get
${({C_6}{H_5}COOH)_n}$
${({C_6}{H_5}COOH)_2}$
It is seen that dimerization of benzoic acid takes place.
The dimerization is the reaction where two similar compounds react to give a new compound.
Thus, the molecular mass of benzoic acid as benzene as determined by depression in the freezing point method corresponds to the dimerization of benzoic acid.

Note:
In the dimerization of benzoic acid in benzene, the Van't Hoff factor is correlated to the degree of association of the acid as shown below.