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:
\[0 - (T_f)_s = 1.86 \times 1\]
\[(T_f)_s = -1.86 \approx -31^\circ \text{C}\]