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

`chi_(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 ethnol in `1 L` of the solution which is nearly `=1 L` of water (because solution is dilute)
Number of moles in `1 L` of water `=(1000 g)/(18 g mol^(-1))`
`=55.55` moles
Substituting `n(H_(2)O)=55.55` in equation (i) we get
`(n(C_(2)H_(5)OH))/(n(C_(2)H_(5)OH)+55.55)=0.040`
or `0.96 n(C_(2)H_(5)OH)=55.55xx0.040`
or `n(C_(2)H_(5)OH)=2.31`
Hence, molarity of the solution`=2.31 M`