InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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. |
|
Answer» Correct Answer - `C_(5)H_(10)O_(5)` `C_xH_(2y)O_y+2xO_2 hArrxCO_2+yH_2O+xO_2` After cooling only `CO_(2)` and `O_(2)` are present because water is in liquid form. Thus, moles of gases after cooling =`x + x =2x` Volume of gases after cooling =`2.24 L` `2x=2.24 L` `x=1.12 L` Number of moles of `CO_(2)=1.12/2.24 =0.5 mol` Hence, the empirical formula of organic compound is `C(H_(2)O)` Given, Vapour pressure of pure water `(P^(@))=17.5 mm Hg` Lowering of vapour pressure `(P^(@)-P)=0.104 mm Hg` weight of solute `(W_(2)) =50 g` weight of solvent `(W_(1)) =1000 g` `(P^(@)-P)/ P^(@)=(W_(2) xx Mw_(1))/(Mw_(1) xx W_(1))` `0.104/17.5=(50 xx 18)/(Mw_(2) xx 1000)` `Mw_(2)=(900 xx 17.5)/104=151.4 g` Empirical formula weight =`C(H_(2)O)` =`12 +2 +16 =30` Molecular weight (m)=151.4 g` `n=("Molecular weight")/("Empirical formula weight")` =`151.4/30 =5.04 =5` Thus, molecular formula =`[C(H_(2)O]_(5)=C_(5)H_(10)O_(5)` |
|