Question

When 2 g of benzoic acid is dissolved in 25 g of benzene, it shows a depression in freezing point equal to 1.62 K. The molal depression constant (K_f) of benzene is 4.9 K kg mol^-1, and the molecular weight of benzoic acid is 122 g/mol. What will be the percentage association of the benzoic acid?

(Benzoic acid forms dimers when dissolved in benzene.)

Answer 1

Weight of solute (w₂) = 2 g

Weight of solvent (w₁) = 25 g

K_f = 4.9 kg mol^-1

ΔT_f = 1.62 K

Now, `Delta"T"_f = "K"_f xx w_2/"M"_1 xx 1000/w_1`

M = `(4.9 xx 2 xx 1000)/(162 xx 25)`

M = 241.98 g/mol

\[\ce{2C6H5COOH<=>(C6H5COOH)2}\]

If x is the degree of association, (1 - x) mole of benzoic acid left undissociated and corresponding `x/2` as associated moles of \[\ce{C6H5COOH}\] at equilibrium.

∴ Total number of moles of particles at the equation

`= 1 - x + x/2 = 1 - x/2` = i

`=> i = "Normal molecular mass"/"Abnormal molecular mass"`

`=> 1 - x/2 = 122/(241.98)`

`=> x/2 = (1 - 122)/241.98`

⇒ x = 0.992

⇒ x = 99.2%

Degree of association of benzoic acid in benzene = 99.2%.