Question

Molecules of benzoic acid $({{C}_{6}}{{H}_{5}}COOH)$ dimerises in benzene. ‘$w$’ g of the acid dissolved in 30g of benzene shows a depression in freezing point equal to 2K. If the percentage association of the acid to form a dimer in solution is 80, then w is: (Given that ${{K}_{f}}$=5K kg/mol Molar mass of benzoic acid=122 g/mol
(A) 1.8g
(B) 2.4g
(C) 1.0g
(D) 1.5g

Answer 1

Hint: Benzoic acid dimerises in benzene due to hydrogen bonding, so Observed molar mass of benzene will be twice of theoretical molar mass, so $\Delta {{T}_{f}}=i\text{ }{{\text{K}}_{f}}m$
If $x$is considered as degree of association, \[i=1-x+\dfrac{x}{2}\]

Complete step by step solution:
-As we know, ionic solid dissociates into cations and anion when dissolved in water.
-When we dissolve one mole of Hydrochloric acid, it is expected that one mole of hydrogen ion and one mole of chloride ions will be released in solution, so then there are two moles of particles present in solution.
-As depression in freezing point is a colligative property which means it depends on the number of particles present in the solution.
-if 1 mole of hydrochloric acid is dissolved in 1 Kg of water, as the number of moles of particles are doubled, so depression in freezing point will be also twice and as we know depression in freezing point is inversely proportional to the molar mass of solute, the observed molecular mass will be 18.25g which is half of the Theoretical molecular mass of Hydrochloric acid 36.5g
Similarly, Benzoic acid dimer in benzene due to hydrogen bonding, so Observed molar mass of benzene will be twice of theoretical molar mass.
Such mass higher or lower than theoretical mass is known as abnormal molar mass.
So $i$ Van’t Hoff factor is the ratio of observed colligative property to calculated colligative property.
So, depression in freezing point for benzoic acid will be given as
\[\Delta {{T }_{f}}=i\text{ }{{\text{K}}_{f}}m\]
Consider benzoic acid at equilibrium,
$2{{C}_{6}}{{H}_{5}}COOH{{({{C}_{6}}{{H}_{5}}COOH)}_{2}}$
If $x$s degree of association, then (1-$x$) mol will be unassociated and $\dfrac{x}{2}$ moles of benzoic acid will be associated.
$i=1-x+\dfrac{x}{2}$
As degree of dissociation is 80%, then
\[i=1-0.8+\dfrac{0.8}{2}=0.6\]
So,
$\begin{align}
& \Delta {{T}_{f}}=i\text{ }{{\text{K}}_{f}}m \\
& 2=0.6\times 5\times \dfrac{w\times 1000}{122\times 30} \\
& \\
\end{align}$

Hence the answer is option (B).

Note: Depression in freezing point is a colligative property which means it depends on the number of particles present in the solution. Benzoic acid dimerises in benzene due to hydrogen bonding, so Observed molar mass of benzene will be twice of theoretical molar mass. Such mass higher or lower than theoretical mass is known as abnormal molar mass. $i$ Van’t Hoff factor is the ratio of observed colligative property to calculated colligative property.