Question

Molecules of benzoic acid (C₆H₅COOH) dimerise 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 dimer in the solution is 80, then w is 

(Given that k_f = 5 K kg mol^-1, molar mass of benzoic acid = 122 g mol^-1
1. 1.8 g
2. 1.0 g
3. 2.4 g
4. 1.5 g

Answer 1

Correct Answer - Option 3 : 2.4 g

Calculation:

Molecules of benzoic acid dimerise in benzene as:

2(C₆H₅COOH) ⇌ (C₆H₅COOH)₂

Now, we know that depression in freezing point (∆T_f) is given by following equation:

\(\Delta {T_f} = i \times {K_f} \times m = \frac{{i \times {K_f} \times {w_{{\rm{solute\;}}}} \times 1000}}{{M{w_{{\rm{solute\;}}}} \times {w_{{\rm{solvent\;}}}}}}\)    ----(1)

Given, w_solute (benzoic acid) = wg

w_solvent (benzene) = 30g

Mw_solute (benzoic acid) = 122gmol^-1, ΔT_f = 2K

K_f = 5 Kkg mol^-1, α = 80 or α = 0.8

2(C₆H₅COOH) ⇌ (C₆H₅COOH)₂

Initial 1 0

Final 1 – α α/2

=1-0.8 0. 8/2=0.4

= 0.2

Total number of moles at equilibrium

= 0.2 + 0.4 = 0.6

\(i = \frac{{{\rm{\;Number\;of\;moles\;at\;equilibrium\;}}}}{{{\rm{\;Number\;of\;moles\;present\;initially\;}}}}\)

\(i = \frac{{0.6}}{1} = 0.6\)

On substituting all the given values in Eq.(1), we get

\(2 = \frac{{0.6 \times 5 \times w \times 1000}}{{122 \times 30}}\) , w = 2.44 g

Thus, weight of acid (w) is 2.4 g.