Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is `0.040`.

Explanation

mole fraction of `C_(2)H_(5)OH =("Number of moles of "H_(2)H_(5)OH)/( "Number of moles of solution ")`
`0.040=(n_(c_(2)H_(5)OH))/(n_(C_(2)H_(5)OH))+n_(H_(2)O)…………………(1)`
Number of moles present in 1 L water :
`n_(H_(2)O)=(1000 g)/(18 h mol ^(-1))`
`n_(H_(2)O)=(55.55 mol`
Subsiiting th value of ` n_(H_(2)O)` in equation (1),
`(n_(C_(2)H_(5)OH))/(n_(C_(2)H_(5)OH)+55.55)=0.040 `
`N_(C_(2)H_(5)OH))=0.040 +(0.040)(55.55) `
`0.96 n_(C_(2)H_(5)OH)+2.222 mol`
`n_(C_(2)H_(5)OH=(2.222)/(0.96) mol`
`n_(C_(2)H_(5)OH) =2.314 mol`
`=(2.314 mol)/(1 L)`
`therefore` Molarity of solution
`=2.14 M`