To solve the problem, we'll calculate the freezing point depression of the acetic acid solution in water.

1. Calculating Moles of Acetic Acid:

• Given volume of acetic acid: $5 \, \text{mL}$


• Density of acetic acid: $1.2 \, \text{g/mL}$


• Molar mass of acetic acid: $60 \, \text{g/mol}$

$\text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g}$

$\text{Moles of acetic acid} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol}$

1. Calculating Molality:

• Volume of water: $1 \, \text{L}$


• Density of water: $1 \, \text{g/cm}^3 \Rightarrow 1 \, \text{kg/L}$


• Hence, mass of water = $1 \, \text{kg}$

$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg}$

1. Degree of Dissociation ($\alpha$):

• Dissociation constant ($K_a$) of acetic acid: $6.25 \times 10^{-5}$

$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 25 \times 10^{-3} = 0.025$

1. Van't Hoff factor (i):

• Acetic acid partially dissociates into 2 ions (CH$_3$COO$^-$ and H$^+$)

$i = 1 + (\text{number of ions produced} - 1) \times \alpha = 1 + (2-1) \times 0.025 = 1 + 0.025 = 1.025$

1. Freezing Point Depression ($\Delta T_f$):

• Freezing point depression constant ($K_f$) for water: $1.86 \,\text{K kg/mol}$

$\Delta T_f = i \times K_f \times \text{Molality} = 1.025 \times 1.86 \times 0.1 = 0.19065 \, \text{K}$

1. Final Freezing Point:

• Pure water freezes at $0 \, ^\circ \text{C}$

$\text{Freezing point of solution} = 0 \, ^\circ \text{C} - 0.19065 \, ^\circ \text{C} = -0.19065 \, ^\circ \text{C}$

Here, $-x \times 10^{-2} \, ^\circ \text{C}$ is given. Therefore, $x = 19$ (nearest integer).

So, the final answer is:

$x = 19$