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

Complete conbustion of organic compound is as follows:
`C_(x)H_(2y)O_(y)+xO_2toxCO_2+yH_2O`
Since oxygen taken is `2xLxLO_2` is left at STP after reaction. Also, `xL` of `CO_2` is formed by 1" mol of "organic compound.
So, `2x=2.24LCO_2`
`thereforex=1.12LCO_2`
or `x=(1.12)/(22.4)molCO_2=0.05molCO_2`
Moles of `H_2O` formed `(y)=(0.9)/(18)=0.05`
`thereforex:y=0.05:0.05=1:1`
`thereforex=1` and `y=1`
`therefore` Empirical formula of organic compound `=30`
Now, molecular weight of compound is deliverd by Raouly's law
`(p^@-P_S)/(p^(@))=(w)/(m)xx(M)/(W)`
`thereforep^@-P_S=` lowering of vapour pressure`=0.104mm`
and `p^@=` vapour pressure of pure solvent `=17.5mm`
`therefore(0.104)/(17.5)=(50)/(m)xx(18)/(1000)`
`thereforem=150.5`
Now, `n=("Molecular weight")/("Empirical formula weight")`
`=(150.5)/(30)=5`
Molecular formala`=` (Empirical formula)`xxn`
`=(CH_2O)xx5=C_5H_(10)O_(5)`