Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.020.

Explanation

`x_(C_(2)H_(5)OH)=(n(C_(2)H_(5)OH))/(n(C_(2)H_(5)OH)+n(H_(2)O))=0.040" (Given) …(i)"`
The aim is to find number of moles of ethanol in 1 L of the solution which is nearly = 1 L of water
(because solution is dilute)
`"No. of moles in 1 L of water"=(1000g)/(18gmol^(-1))="55.55 moles"`
substituting n `(H_(2)O)="55.55 in eqn (i), we get "`
`(n(C_(2)H_(4)OH))/(n(C_(2)H_(5)OH)+55.55)=0.040 or 0.96n(C_(2)H_(5)OH)=55.55xx0.040or n(C_(2)H_(5)OH)="2.51 mol"`
Hence, molarity of the solution = 2.31 M