Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).

Explanation

Number of moles of ethanol present in 1 L of aqueous solution indicates the molarity of the solution. Let 1L water be present in 1L ethanol solution [since, the solution is dilute].
Number of moles of water present in 1 L water
`n_(H_(2)O) =(1000g)/(18g*mol^(-1))=55.55` mol
For binary solutions, mole fraction of the first component+ Mole fraction of the second component=1.
therefore, in this case `x_(H_(2)O)+x_(C_(2)H_(5)OH)=1`
or, `x_(H_(2)O)=1-x_(C_(2)H_(5)OH) or, x_(H_(2)O)=1-0.040=0.96`
Again, `x_(H_(2)O)=(n)/(n_(H_(2)O)+n_(C_(2)H_(5)OH)) or, 0.96=(55.55)/(55.55+n_(C_(2)H_(5)OH))`
or, `53.328+0.96n_(C_(2)H_(5)OH) =55.55`
`therefore n_(C_(2)H_(5)OH) =(2.222)/(0.96)=2.3146` mol
`therefore` Molarity of the solution`=2.3146mol*L^(-1)`.