Question

The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g 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 = 5.12 K kg mol⁻¹)

विकल्प

• 74.6%

• 94.6%

• 64.6%

• 80.4%

Answer 1

The freezing point of benzene decreases by 0.45°C when 0.2 g of acetic acid is added to 20 g of benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be 94.6%.

Explanation:

Given: Mass of solute (acetic acid) = 0.2 g

Molar mass of acetic acid (CH₃COOH) = 60 g/mol

Mass of benzene = 20 g = 0.020 kg

ΔT_f = 0.45°C

K_f = 5.12 K kg mol⁻¹

Acetic acid associates to form a dimer, so we must find the percentage association.

Moles of acetic acid = `0.2/60`

= 0.00333 mol

Molality = `0.00333/0.020`

= 0.1667 mol/kg

ΔT_f = i K_f m

`i = (Delta T_f)/(K_f * m)`

= `0.45/(5.12 xx 0.1667)`

i = 0.526

For dimerisation

\[\ce{2A <=> A2}\]

`i = 1 - alpha/2`

⇒ `0.526 = 1 - alpha/2`

⇒ `alpha/2 = 1 - 0.526`

= 0.474

α = 0.948

Percentage association = α × 100

= 0.948 × 100

= 94.8%