A chloro compound "A".
(i) forms aldehydes on ozonolysis followed by the hydrolysis.
(ii) when vaporized completely 1.53 g of A, gives 448 mL of vapour at STP. The number of carbon atoms in a molecule of compound A is _________

We are told that compound $$A$$ is a chloro-compound and that on ozonolysis followed by hydrolysis it produces only aldehydes. Recall that:

Whenever an alkene $$\,{\displaystyle R_1CH=CHR_2}\,$$ is treated with ozone and then hydrolysed, the double bond is split and each carbon of the double bond becomes the carbon of a carbonyl group. If every carbon of the original double bond possessed at least one hydrogen, both carbonyl products will be aldehydes. Hence we infer that $$A$$ is an alkene that contains one chlorine atom and possesses exactly one C=C double bond.

Let its molecular formula be written as $$\mathrm{C_nH_mCl}$$. We next determine its molar mass from the vapour-density data.

At STP, one mole of any gas occupies $$22.4\ \text{L}$$. Here,

Volume of vapour obtained  =  $$448\ \text{mL}=0.448\ \text{L}$$.

So, number of moles present is

$$n_{\text{mole}}=\dfrac{\text{volume}}{22.4\ \text{L mol}^{-1}} =\dfrac{0.448}{22.4}=0.020\ \text{mol}.$$

The given mass that produced this vapour is $$1.53\ \text{g}$$, hence the molar mass $$M$$ of $$A$$ is

$$M=\dfrac{\text{mass}}{\text{moles}} =\dfrac{1.53\ \text{g}}{0.020\ \text{mol}} =76.5\ \text{g mol}^{-1}.$$

Writing the molar mass in terms of the unknown numbers of atoms, we have

$$M = 12n + m + 35.5,$$

because one carbon contributes $$12\ \text{u}$$, one hydrogen gives $$1\ \text{u}$$, and one chlorine contributes $$35.5\ \text{u}$$ to the molar mass.

Equating this to the experimentally obtained value,

$$12n + m + 35.5 = 76.5 \quad\Longrightarrow\quad 12n + m = 41.$$

Next, we employ the idea of the degree of unsaturation (also called the index of hydrogen deficiency, IHD). For a straight-chain saturated chloro-alkane with $$n$$ carbons we would expect $$2n+2$$ hydrogens. Each ring, double bond or halogen atom reduces this count by the following rule:

IHD formula: $$\text{IHD}=\dfrac{2n+2 - (m + X)}{2},$$ where $$X$$ is the number of halogen atoms.

In our case there is one Cl atom, so $$X=1$$. Because we have already concluded that there is one C=C double bond and no rings, $$\text{IHD}=1$$. Therefore

$$1=\dfrac{2n+2 - (m+1)}{2} \;\Longrightarrow\; 2=2n+2 - m -1 \;\Longrightarrow\; m = 2n -1.$$

We now possess two simultaneous relations, namely

$$12n + m = 41,$$

$$m = 2n - 1.$$

Substituting the second into the first,

$$12n + (2n - 1) = 41 \;\Longrightarrow\; 14n - 1 = 41 \;\Longrightarrow\; 14n = 42 \;\Longrightarrow\; n = 3.$$

Thus the molecule contains exactly $$3$$ carbon atoms. Immediately we can verify the hydrogen count:

$$m = 2n - 1 = 2(3) - 1 = 5,$$

which gives the molecular formula $$\mathrm{C_3H_5Cl}$$ and satisfies the molar mass check:

$$M = 12(3) + 5 + 35.5 = 36 + 5 + 35.5 = 76.5\ \text{g mol}^{-1},$$

in full agreement with the experimental value.

Hence, the correct answer is Option 3.