1. Find moles of acetic acid added

Given:

• Volume of acetic acid $=5\,\text{mL}$
• Density of acetic acid $=1.2\,\text{g mL}^{-1}$

So, mass of acetic acid: $m=5\times 1.2=6\,\text{g}$

Moles of acetic acid: $n=\frac{6}{60}=0.1\,\text{mol}$

2. Volume and mass of water (solvent)

Given $1$ litre water. Since density of water $=1\,\text{g cm}^{-3}$, $1\,\text{L}=1000\,\text{mL}=1000\,\text{g}=1\,\text{kg}$

Thus mass of solvent $=1\,\text{kg}$.

3. Initial concentration of acetic acid

Since total volume is approximately $1\,\text{L}$, the molarity is $C\approx \frac{0.1}{1}=0.1\,\text{M}$

For dissociation: $\mathrm{CH_3COOH} \rightleftharpoons \mathrm{CH_3COO^-}+\mathrm{H^+}$

Let degree of dissociation be $\alpha$. Then, $K_a=\frac{C\alpha^2}{1-\alpha}$

Given: $K_a=6.25\times 10^{-5},\quad C=0.1$

Since acetic acid is weak, $\alpha\ll 1$, so $1-\alpha\approx 1$. Hence, $K_a\approx C\alpha^2$

So, $\alpha=\sqrt{\frac{K_a}{C}}=\sqrt{\frac{6.25\times 10^{-5}}{0.1}}$ $=\sqrt{6.25\times 10^{-4}}=2.5\times 10^{-2}=0.025$

4. Van’t Hoff factor

Acetic acid gives $2$ particles on dissociation, so $i=1+\alpha=1.025$

5. Molality of solution

Moles of solute $=0.1$ mol, mass of solvent $=1$ kg, $m=\frac{0.1}{1}=0.1\,\text{mol kg}^{-1}$

6. Depression in freezing point

$\Delta T_f=iK_f m$

Substitute values: $\Delta T_f=1.025\times 1.86\times 0.1$ $=0.19065\,\text{K}$

Thus freezing point is lowered by about $0.19065^\circ\text{C}=19.065\times 10^{-2}\,^\circ\text{C}$

So, $x\approx 19$

7. Final answer

The solution freezes at $-19\times 10^{-2}\,^\circ\text{C}$ so the required integer is $\boxed{19}$