A solute HA (formal molality $0.20 mol k g^{- 1}$ ) dimerises in benzene ( $K_{f} = 5.12 K kg mo l^{- 1}$ ), and $80%$ of the HA is present as dimers. Which statements are correct?

AThe van't Hoff factor is $0.60$
BThe molar mass calculated from the freezing point depression is about $1.67$ times the true molar mass of HA
CThe observed depression is $0.819 K$
DThe observed depression is larger than it would be if there were no association

Answer: A, B

i = 1 - α /2 = 1 - 0.40 = 0.60

. Observed

Δ T_{f} = i K_{f} m = 0.60 \times 5.12 \times 0.20 = 0.614 K

, which is smaller than the value without association (

1.024 K

); the value

0.819

would need

i = 0.80

. Because the depression is lower, the apparent molar mass is higher by

1/ i = 1.67

times.