1. Find molar mass of compound $A$ from vapour data

At STP, $1$ mole of gas occupies $22.4\,\text{L} = 22400\,\text{mL}$.

Given volume of vapour: $448\,\text{mL} = 0.448\,\text{L}$

So moles of $A$ are: $n = \frac{0.448}{22.4} = 0.02\,\text{mol}$

Given mass of $A$ is $1.53\,\text{g}$, hence molar mass: $M = \frac{1.53}{0.02} = 76.5\,\text{g mol}^{-1}$

2. Use the fact that $A$ is a chloro compound

Let the molecular formula of a monochloro compound be: $\mathrm{C_nH_mCl}$

Its molar mass is: $12n + m + 35.5 = 76.5$

So, $12n + m = 41$

Now test integer values of $n$:

• If $n=1$, then $m=29$ impossible
• If $n=2$, then $m=17$ impossible
• If $n=3$, then $m=5$ possible, giving formula: $\mathrm{C_3H_5Cl}$
• If $n=4$, then $m=-7$ impossible

Thus the molecular formula is: $\mathrm{C_3H_5Cl}$

So the compound has 3 carbon atoms.

3. Check ozonolysis condition

Degree of unsaturation for $\mathrm{C_3H_5Cl}$:

For a saturated acyclic monochloro compound, formula would be $\mathrm{C_nH_{2n+1}Cl}$. For $n=3$, saturated formula is $\mathrm{C_3H_7Cl}$. But actual formula is $\mathrm{C_3H_5Cl}$, i.e. it has two fewer H, so it contains one double bond.

Hence $A$ is an alkenyl chloride.

A suitable structure is allyl chloride: $\mathrm{CH_2=CH-CH_2Cl}$

On ozonolysis of the double bond followed by hydrolysis, it forms aldehyde-containing products (notably formaldehyde), which is consistent with the statement.

Thus the data are self-consistent.

4. Final answer

The number of carbon atoms in compound $A$ is: $\boxed{3}$