An organic compound `(C_(x)H_(2y)O_(y))` was burnt with twice the amount of oxygen needed for complete combustion to `CO_(2)` and `H_(2)O`. The hot gases, when cooled to `0^(@)C` and `1 atm` pressure, measured `2.24 L`. The water collected during cooling weighed `0.9 g`. The vapour pressure of pure water at `20^(@)C` is `17.5 mm Hg` and is lowered by `0.104 mm` when `50 g` of the organic compound is dissolved in `1000 g` of water. Give the molecular formula of the organic compound.

Explanation

To find the molecular formula of the organic compound \( C_xH_{2y}O_y \), we will follow these steps:

### Step 1: Determine the moles of gases produced
The volume of the gases measured after combustion is given as 2.24 L at standard temperature and pressure (STP). At STP, 1 mole of gas occupies 22.4 L.

\[ \text{Moles of gas} = \frac{\text{Volume}}{\text{Molar volume}} = \frac{2.24 \, \text{L}}{22.4 \, \text{L/mol}} = 0.1 \, \text{mol} \]
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