A chemist prepared a solution by adding by `230 g` of pure ethanol `(C_(2)H_(5)OH)` to `144 g` of water. Calculate the mole fractions of these two components. The molar masses of ethanol and water are `64 g` and `18 g`, respectively.
Strategy: To calculate the mole fractions, we must find the number of moles of ethanol and of water. To find the moles, we must divide the given mass by the respective molar mass.
In a two-component system made up of `A` and `B` molecules, the mole fraction of a component of a solution, say, component A is written `chi_(A)` and is defined as
Mole fraction of component `A= chi_(A)`
`=("moles of A")/("sum of moles of A and B")`

Explanation

The number of moles of `C_(2)H_(5)OH and H_(2)O` present in solution are
Moles of ethnol `(n_(C_(2)H_(5)OH))= ("mass of ethanol")/("molar mass of ethanol")`
`=(230 g C_(2)H_(5)OH)/(46 g C_(2)H_(5)OH//"mole" C_(2)H_(5)OH)`
`=5 mol C_(2)H_(5)OH`
Moles of water `(n_(H_(2)O))=("mass of water")/("molar mass of water")`
`=(144 g H_(2)O)/(18 g H_(2)O //"mol" H_(2)O)`
`=8 "mol" H_(2)O`
Using Equation (2.7), we can write the mole fractions of ethanol and water as
`chi_(C_(2)H_(5)OH)=n_(C_(2)H_(5)OH)/(n_(C_(2)H_(5)OH)+n_(H_(2)O))`
`=(5 mol)/((5+8) mol)`
`chi_(H_(2)O)=n_(H_(2)O)/(n_(C_(2)H_(5)OH)+n_(H_(2)O))`
`=(8 mol)/((5+8) mol)=0.6154`