When `2g` o non`-` volatile hydrocarbon containing `94.4` percent carbon is dissolved in `100g` benzene, the vapour pressure of Benzene is lowered from `74.66 ` torr to `74.01` torr. Determine 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)`