Vapor pressure of a solution of Volatile components and composition of vapor: Consider a solution containing `738 g` of water and `253 g` of ethanol `(C_(2)H_(5)OH)` at `323 K`. At this temperature, the vapour pressure of pure ethanol is `0.292 atm` and the vapor pressure of pure water is `0.122 atm`. Calculate the vapour pressure of the solution and mole fraction of every component in vapour phase.
Strategy: First calculate the moles and the mole fraction of each component in the liquid solution. Then apply Raoult's law to each of the two volatile components to calculate their partial pressures. The total vapour pressure is the sum of the partial vapour pressures of the components and the mole fraction of a components in a gaseous mixture equals the ratio of its partial pressure to the total pressure.

Explanation

Step 1. Calculate the moles of each component
Molecular mass of `H_(2)O=1xx2+16=18u`
Molar mass of `H_(2)O=18 g mol^(-1)`
Molecular mass of `C_(2)H_(5)OH =2xx12+6xx1+1xx16`
`=46 u`
Molar mass of `C_(2)H_(5)OH =46 g mol^(-1)`
`n_(H_(2)O)=(mass_(H_(2)O))/(molar mass_(H_(2)O))`
`=(738 g H_(2)O)/(18 g mol^(-1)H_(2)O)`
`=41 mol H_(2)O`
`n_(C_(2)H_(5)OH)=(mass_(C_(2)H_(5)OH))/(molar mass_(C_(2)H_(5)OH))`
`=(253 g C_(2)H_(5)OH)/(46 g mol^(-1) C_(2)H_(5)OH)`
`=5.5 mol C_(2)H_(5)OH`
Total number of moles `(n_("total"))=41 mol H_(2)O+5.5 mol C_(2)H_(5)OH`
`=46.5 mol`
Step 2. Calculate the mole fraction of each component in the liquid solution
`chi_(H_(2)O)=n_(H_(2)O)/(n_(H_(2)O)+n_(C_(2)H_(5)OH))`
`=(41 mol)/(41 mol + 5.5 mol)`
`=0.882`
`chi_(C_(2)H_(5)OH)= 1- chi_(H_(2)O)`
`=1- 0.882=0.118`
Step 3. Apply Raoult's law for volatile components
`P_(H_(2)O)=P_(H_(2)O)^(0)chi_(H_(2)O)`
`=(0.122 atm)(0.882)`
`=0.108 atm`
`P_(C_(2)H_(5)OH)=P_(C_(2)H_(5)OH)^(0)chi_(C_(2)H_(5)OH)`
`=(0.292 atm)(0.118)`
`=0.0344 atm`
`P_("total")=P_(H_(2)O)+P_(C_(2)H_(5)OH)`
`=0.108 atm + 0.0344 atm`
`0.1424 atm`
Step 4. Applying Dalton's law as the vapor above the solution a mixture of ideal gases.
`Y_(H_(2)O(g))=P_(H_(2)O)/P_("total")`
`=(0108 atm)/(0.1424 atm)`
`=0.758`
`Y_(C_(2)H_(5)OH(g))=1- Y_(H_(2)O(g))`
`=1- 0.758`
`=0.242`