Question

The freezing point of benzene decreases by ${0.45^ \circ }C$ when $0.2g$ of acetic acid is added to $20g$ of benzene . If acetic acid associates to form a dimer in benzene , percentage association of acetic acid in benzene will be :
[ ${K_f}$ for benzene = $5.12Kkgmo{l^{ - 1}}$ ]
A.$74.6\%$
B.$94.6\%$
C.$64.6\%$
D.$80.4\%$

Answer 1

When a non - volatile solute is added to a volatile solvent , then vapour pressure decreases , boiling point increases and freezing point decreases . Freezing point of a substance is a colligative property .
Formula used : $\Delta {T_f} = i{K_f}m$

Complete step by step answer:
When acetic acid is dissolved in benzene then association takes place . So , after association , number of solute particles decreases , so colligative property decreases and molecular mass increases .
The dimerization of benzene takes place
$2C{H_3}COOH \rightleftharpoons {(C{H_3}COOH)_2}$
Before the dimerization starts the quantity of acetic acid is 1 mole
After the reaction proceeds the quantity of dimer of acetic acid is $\dfrac{x}{2}$ and $1 - x$ of reactant is left
Hence we can say that number of particles after association are = $1 - x + \dfrac{x}{2} = 1 - \dfrac{x}{2}$
The formula which we will be using in this question is $\Delta {T_f} = i{K_f}m$
where , $\Delta {T_f}$ = depression in freezing point
$i$ = Van't hoff factor
${K_f}$ = molal depression constant
$m$ = molality
First we will take out the molality of the solution
$m = \dfrac{{{w_b} \times 1000}}{{{M_b} \times {w_a}}}$ where , ${w_b}$ = weight of solute , ${M_b}$ = molecular mass of solute and ${w_a}$ = weight of the solvent .
It is given that $0.2g$ of acetic acid is added to $20g$ of benzene ( molecular mass of benzene = $60gmo{l^{ - 1}}$ )
So , on substituting the values in the molality formula , we get
$m = \dfrac{{0.2 \times 1000}}{{60 \times 20}} = 0.1666$
To calculate the Van't Hoff factor the following formula is used
$i = \dfrac{{total\,number\,of\,particles\,after\,association}}{{total\,number\,of\,particles\,before\,association}}$
So , $i = \dfrac{{1 - \dfrac{x}{2}}}{1} = 1 - \dfrac{x}{2}$
So , now we know the values of i and m , also $\Delta {T_f} = 0.45$ and ${K_f} = 5.12Kkgmo{l^{ - 1}}$
So , now on substituting the values in $\Delta {T_f} = i{K_f}m$ , we get
$0.45 = (1 - \dfrac{x}{2})(5.12)(0.1666)$
$\Rightarrow 1 - \dfrac{x}{2} = 0.527$
$\Rightarrow x = 0.946$
Hence the degree of association is $94.6\%$ .
So option B is correct .

Note: The value of ${K_f}$ depends upon the nature of the solvent , that is , different solvents have different values of molal depression constant and its value does not change for a given solute.