`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 ?

Options

A. `C_(6)H_(6)`
B. `C_(12)H_(6)`
C. `C_(7)H_(6)`
D. `C_(14)H_(10)`

Answer

4

Explanation

According to Raoutls law
`(Deltap)/(p^(0))=x_(2)`
Where `-Delta p =(74.01-74.66)` torr and `p^(0)=74.66` torr ltbtgt If M is the molar mass of hydrocarbon, then
`X_(2)=(n_(2))/(n_(1)+n_(2))=((8)/(M))/(((100)/(78))+((2)/(M)))`
Hence `(74.66-74.01)/(74.66)=((2)/(M))/((100)/(78)+(2)/(M))`
Solving for M, we get , `M =177.6 g mol^(-1)` .
Given mass ratio is `m_(c):m_(H): :934.4:5.6`
This atomic ratio is
`N_(C): N_(H): :(84.4)/(12):(5.6)/(1)implies7.87:5.6" "implies1.4:1implies7:5`
Hence, Empirical formula is `C_(7)H_(5)`
Molar Emipirical mass `89g mol^(-1)`
Number of `C_(7)H_(5)` unit in the given molecule `=("Molar mass")/("Molar empirical mass")=(177.6)/(89)-=2`
Thus molecular formula is `C_(14)H_(10)`