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

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