Question

The freezing point depression of 0.1 molal solution of acetic acid in benzene is 0.25 K. K_f for benzene is 5.12 K kg mol⁻¹. What conclusion can you draw about the molecular state of acetic acid in benzene?

Answer 1

Given: Molality (m) = 0.1 mol/kg

Depression in freezing point (ΔT_f) = 0.25 K

K_f for benzene = 5.12 K kg mol⁻¹

If there will be no association, then dissociation.

`Delta_f^"expeccted" = K_f * m`

= 5.12 × 0.1

= 0.512 K

Van’t Hoff factor (i) = `(Delta_f^"observed")/(Delta_f^"expeccted")`

= `0.25/0.512`

Vant Hoff factor (i) = 0.49

Here, i < 1 indicates the association of solute particles and acetic acid molecules in benzene, likely forming dimers due to hydrogen bonding.

Since the observed depression in freezing point is less than expected, the van’t Hoff factor is less than 1, indicating that acetic acid undergoes association (dimerisation) in benzene due to hydrogen bonding.

This behaviour is common for polar molecules like acetic acid in non-polar solvents.