`2g` of a non-volatile hydrocarbon solute dissolved in `100g` of a hypothetical organic solvent (molar mass `=50`) was found to lower the vapour pressure from `75.00` to `74.50 mm` of `Hg` at `20^(@)C`. Given that the hydrocarbon contains `96%` of `C`, what is the molecular formula of the hydrocarbon ?

Answer

`C_(12)H_(6)`

Explanation

`(75-74.50)/(75)=(((2)/(M_(2))))/(((2)/(M_(2)))+(100)/(50))`
where `M_(2)` is the molecular weight of the solute,
`(0.50)/(75)=0.006667=(1)/(1+M_(2))`
so, `M_(2)=149 g//"mole"`
Percentage of Carbon `=C=96%`
`H=4%`
`C:H=2:1`
`(C_(2)H_(1))_(n)=149//25 rArr n=6`
Molecular formula of the hydrocarbon `C_(12)H_(6)`