Saved Bookmarks
| 1. |
`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 ?A. `C_(6)H_(6)`B. `C_(12)H_(6)`C. `C_(7)H_(6)`D. `C_(14)H_(10)` |
|
Answer» Correct Answer - 4 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)` |
|