Question

There is $\text{ }50{\scriptstyle{}^{0}/{}_{0}}\text{ }$ a dimer formation of benzoic acid $\text{ }\left( {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH} \right)\text{ }$ in benzene solution.
\[\text{2}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH}\rightleftharpoons {{\text{(}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH)}}_{\text{2}}}\],
Hence the abnormal molecular weight of benzoic acid (theoretical value\[\text{ 122 g mo}{{\text{l}}^{\text{-1}}}\text{ }\] ) is:
A) \[\text{ 61 g mo}{{\text{l}}^{\text{-1}}}\text{ }\]
B) \[\text{ 244 g mo}{{\text{l}}^{\text{-1}}}\text{ }\]
C) \[\text{ 163 g mo}{{\text{l}}^{\text{-1}}}\text{ }\]
D) \[\text{ 81 g mo}{{\text{l}}^{\text{-1}}}\text{ }\]

Answer 1

Hint: If n simple molecules of the combine to form associated molecule i.e. $\text{ nA }\rightleftharpoons \text{ (A}{{\text{)}}_{\text{n}}}\text{ }$.Then $\text{ }\!\!\alpha\!\!\text{ }$ is the degree of dissociation and the van’t Hoff Factor ‘i’ is related to the abnormal or normal molar mass as:
\[\text{ i = }\dfrac{\text{ Normal molar mass}}{\text{Observed molar mass}}\text{ }\]

Complete step by step answer:
There are many organic solutes which in non-aqueous solutions undergo the association that the two or more molecules of the solute associate to form a bigger molecule.
Here benzoic acid $\text{ }\left( {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH} \right)\text{ }$ undergoes the association of molecules in benzene solution to form a dimer of benzoic acid ${{\left( {{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH} \right)}_{\text{2}}}$.

Let's consider the association of benzoic acid as follows:
\[\begin{align}
  & \begin{matrix}
   {} & \text{2}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH} & \rightleftharpoons & {{\text{(}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}\text{COOH)}}_{\text{2}}} \\
   \text{initial }\!\!~\!\!\text{ } & \text{1} & {} & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ -} \\
   {} & {} & {} & {} \\
   \text{At }\!\!~\!\!\text{ e}{{\text{q}}^{\text{m}}}\text{ }\!\!~\!\!\text{ } & \text{ }\!\!~\!\!\text{ }\!\!~\!\!\text{ 1- }\!\!\alpha\!\!\text{ } & {} & \dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}} \\
\end{matrix} \\
 & \text{ } \\
\end{align}\]
Here $\text{ }\!\!\alpha\!\!\text{ }$ is the degree of Association. By degree of dissociation is meant the fraction of the total number of molecules that combine to form a bigger molecule.
It $\text{ }\!\!\alpha\!\!\text{ }$ is the degree of association. Then,
The number of unassociated moles equals to $\text{1- }\!\!\alpha\!\!\text{ }$
The number of associated moles is equal to \[\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}\]

Therefore, the number of effective moles is $\text{ 1- }\!\!\alpha\!\!\text{ + }\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}\text{ }$
We are given that the benzoic acid undergoes the $\text{ }50\text{ }{\scriptstyle{}^{0}/{}_{0}}$ dimerization. Thus the ‘n’ stands for the number of molecules that are associated to form a bigger molecule. Therefore, $\text{n=2}$

Thus, the degree of dissociation can be calculated as:
\[\begin{align}
  & \dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}=0.5 \\
 & \Rightarrow \text{ }\!\!\alpha\!\!\text{ = 0}\text{.5 }\times \text{ 2 = 1}\text{.0} \\
 & \text{(since n = 2 )} \\
\end{align}\]
We know that the van’t Hoff factor is used in the account of the abnormal cases. It is denoted by the term ‘i’. Thus van’t Hoff factor for abnormal mass is given as:
\[\text{ i = }\dfrac{\text{ Normal molar mass}}{\text{Observed molar mass}}\text{ }\] (1)

Since we know that the van’t Hoff factor is proportional to the number of moles, therefore ‘i’ is given by:
\[\text{ i = }\dfrac{\text{ Observed osmotic effect}}{\text{Calculated osmotic effect}}\text{ = }\dfrac{\text{1- }\!\!\alpha\!\!\text{ + }\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}}{\text{1}}\text{ }\] (2)
On equating (1) and (2) we get,
\[\text{ i = }\dfrac{\text{ Normal molar mass}}{\text{Observed molar mass}}\text{ = }\dfrac{\text{1- }\!\!\alpha\!\!\text{ + }\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}}{\text{1}}\text{ }\] (3)

Let us substitute the values in equation (3) we get
\[\text{ i = }\dfrac{\text{ Normal molar mass}}{\text{Observed molar mass}}\text{ = }\dfrac{\text{1- }\!\!\alpha\!\!\text{ + }\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{n}}}{\text{1}}\text{ }\]
Or \[\text{ i = }\dfrac{\text{ 122 g mo}{{\text{l}}^{\text{-1}}}}{\text{Observed molar mass}}\text{ = }\dfrac{\text{1- 1 + }\dfrac{1}{\text{2}}}{\text{1}}\text{ }\]
Or \[\text{ i = }\dfrac{\text{ 122 g mo}{{\text{l}}^{\text{-1}}}}{\text{Observed molar mass}}\text{ = }\dfrac{\text{1- 1 + }\dfrac{1}{\text{2}}}{\text{1}}\text{ }\]
Or $\text{ Observed molar mass = 122 }\!\!\times\!\!\text{ 2 g mo}{{\text{l}}^{\text{-1}}}$

Thus, the observed or abnormal molar mass of the benzoic acid is $244\text{ g mo}{{\text{l}}^{\text{-1}}}$
So, the correct answer is “Option B”.

Note: The observed molar mass of the compound is calculated by using the colligative properties. For example, For elevation in boiling point, the molar mass is calculated as:

\[\text{ }{{\text{M}}_{\text{2}}}\text{ = }\left( \dfrac{{{\text{K}}_{\text{b}}}\text{ }{{\text{w}}_{\text{2}}}}{{{\text{w}}_{\text{1}}}\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}} \right)\text{ }\] Where, \[{{\text{w}}_{\text{1}}}\] is the mass of the solvent.