Given : $${C_x}{H_{2y}}{O_y} + 2x{O_2} \to xC{O_2} + y{H_2}O + x{O_2}$$

After cooling, only CO₂ and O₂ will be present in gaseous state because water exists in liquid form at 0$$^\circ$$C and 1 atm.

$$\therefore$$ Number of moles of gases after cooling = x + x = 2x

Volume of gases after cooling = 2.24 L (given)

$$\therefore$$ Moles of gases after cooling $$ = {{2.24} \over {22.4}} = 0.1$$

$$\therefore$$ 0.1 = 2x, or, x = 0.05

As given in the question, water collected during cooling = 0.9 g

$$\therefore$$ Number of moles of water $$ = {{0.9} \over {18}} = 0.05$$; hence, y = 0.05

$$\therefore$$ The empirical formula of the organic compound is CH₂O.

Given, vapour pressure of pure water (p⁰) = 17.5 mm Hg

Lowering of vapour pressure (p⁰ $$-$$ p) = 0.104 mm Hg

No. of moles of water in solution $$({n_1}) = {{1000} \over {18}} = 55.55$$ mol of the molar mass of solute is Mg mol^$$-$$1, then the number of moles of solute in solution $$({n_2}) = {{50} \over M}$$ mol

According to Raoults' law, $${{{p^0} - p} \over {{p^0}}} = {x_2} \approx {{{n_2}} \over {{n_1}}}$$

$$\therefore$$ $${{0.104} \over {17.5}} = {{50} \over {M \times 55.55}}$$ or, $$M = {{50 \times 17.5} \over {55.55 \times 0.104}}$$ or, M = 151.4

$$\therefore$$ Molar mass of the organic compound = 151.4 g mol^$$-$$1

Now, empirical formula mass of CH₂O = 30 g mol^$$-$$1

$$\therefore$$ $$n = {{Molecular\,mass} \over {Empirical\,formula\,mass}} = {{151.4} \over {30}}$$ or, $$n = 5.04 \simeq 5$$

$$\therefore$$ Molecular formula $$ = {[C({H_2}O)]_5} = {C_5}{H_{10}}{O_5}$$