An organic compound `C_xH_(2y)O_y` was burnt with twice the amount of oxygen needed for complete combustion to `CO_2` and `H_2O`. The hot gases when cooled to `0^@C` and 1 atm pressure, measured `2.24` litre. The water collected during cooling weighed 0.9 gm. the vapour pressure of pure water at `20^@C` is 17.5 mm Hg and is lowered by 0.104 mm when 50 gm of the organic compound is dissolved in 1000 gm of water. Give the molecular formula of the organic compound.

Explanation

Complete combustion of organic compound is as:
`C_xH_(2y)O_yrarrxCO_2+yH_2O`
Since oxygen taken is `2x` litres `x` litres `x` litres `O_2` is left at STP after reaction. Also `x` litre of `CO_4` is formed by 1 mol of organic compound.
So, `2x=2.25" litre "CO_2`
`x=1.12" litre "CO_2`
or `x=(1.12)/(22.4)molCO_2`
`=0.05molCO_2`
Moles of `H_2O` formed `(y)=(0.9)/(18)=0.05`
`x:y=0.05:0.05=1:1`
`x=1` and `y=1`
Empirical formula of organic compound `=CH_2O`
Empirical formula weight of organic compound `=30`
Now, molecular weight of compound is derived by Raoult's law:
`(p^@-p_S)/(p^@)=(w)/(m)xx(M)/(W)`
`p^@-P_S=` lowering of vapour pressure `=0.104mm`
and `p^@=` vapour pressure of pure solvent `=17.5mm`
`(0.104)/(17.5)=(50)/(m)xx(18)/(1000)`
`m=150.5`
Now, `n=("molecular weight")/("empirical formula weight")`
`=(150.5)/(30)=5`
Molecular formula`=("empirical formula")xxn`
`=(CH_2O)xx5=C_5H_(10)O_5`