Question

2.7 Kg of each of water and acetic acid are mixed. The freezing point of the solution will be –x °C. Consider the acetic acid does not dimerise in water, nor dissociates in water. \( x = \) ______ (nearest integer).
[Given: Molar mass of water = \( 18 \, \text{g mol}^{-1} \), acetic acid = \( 60 \, \text{g mol}^{-1} \) \( K_f \, \text{H}_2\text{O} = 1.86 \, \text{K kg mol}^{-1} \) \( K_f \, \text{acetic acid} = 3.90 \, \text{K kg mol}^{-1} \) Freezing point: \( \text{H}_2\text{O} = 273 \, \text{K}, \, \text{acetic acid} = 290 \, \text{K} \)]

Answer 1

Correct Answer: 31

Since the moles of water are greater than the moles of \( \text{CH}_3\text{COOH} \), water acts as the solvent. The freezing point depression is calculated as:
\[T_f^0 - (T_f)_s = K_f \times m\]
Calculating the molality \( m \):
\[m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{2700/60}{2700/1000} = 1 \, \text{mol kg}^{-1}\]
Applying the freezing point depression formula:
\[0 - (T_f)_s = 1.86 \times 1\]
\[(T_f)_s = -1.86 \approx -31^\circ \text{C}\]

Answer 2

Step 1: Parse the data and the goal
We have a binary liquid mixture formed by water and acetic acid, each of mass 2.7 kg. Neither dimerisation nor dissociation occurs. We are asked for the freezing point of the mixture in the form −x °C, i.e., find x (nearest integer).

Given constants:
• Molar mass: water = 18 g mol⁻¹, acetic acid = 60 g mol⁻¹
• Kf (water) = 1.86 K kg mol⁻¹, Kf (acetic acid) = 3.90 K kg mol⁻¹
• Pure freezing points: water = 273 K, acetic acid = 290 K

Step 2: Strategy for a two-solvent system
In a two-solvent mixture, each component can act as the solvent while the other acts as the solute. The mixture will begin to freeze when the first component reaches its depressed freezing point. Therefore:
1) Compute the depressed freezing point of water when acetic acid is the solute.
2) Compute the depressed freezing point of acetic acid when water is the solute.
3) The mixture starts to freeze at the higher (less depressed) of these two temperatures, because that component crystallises first as we cool.

Step 3: Depression of freezing point of water (acetic acid as solute)
Moles of acetic acid = 2700 g ÷ 60 g mol⁻¹ = 45 mol.
Mass of water (solvent) = 2.7 kg ⇒ molality of acetic acid in water:
m(acid in water) = 45 mol ÷ 2.7 kg = 16.666... mol kg⁻¹.
Freezing point depression of water:
ΔT_f(water) = Kf(water) × m = 1.86 × 16.666... = 31.0 K.
Depressed freezing point of water:
T_f,water = 273 K − 31.0 K = 242 K = 242 − 273 = −31 °C.

Step 4: Depression of freezing point of acetic acid (water as solute)
Moles of water = 2700 g ÷ 18 g mol⁻¹ = 150 mol.
Mass of acetic acid (solvent) = 2.7 kg ⇒ molality of water in acetic acid:
m(water in acid) = 150 mol ÷ 2.7 kg = 55.555... mol kg⁻¹.
Freezing point depression of acetic acid:
ΔT_f(acid) = Kf(acid) × m = 3.90 × 55.555... = 216.7 K (approx).
Depressed freezing point of acetic acid:
T_f,acid = 290 K − 216.7 K ≈ 73.3 K = 73.3 − 273 = −199.7 °C (very low).

Step 5: Select the actual freezing point of the mixture
As the mixture cools, the component that reaches its (depressed) freezing point first crystallises. Here, water reaches −31 °C, whereas acetic acid would not freeze until about −200 °C. Therefore, the mixture begins to freeze at the higher temperature, i.e., at −31 °C (due to water crystallisation).

Step 6: Report x (nearest integer)
The freezing point is −x °C = −31 °C ⇒

31