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

Explanation

Mole fraction of ethanol
`x(C_(2)H_(5)OH) = n(C_(2)H_(5)OH)/(n(C_(2)H_(5)OH) + n(H_(2)O))`
To calculate molarity, we need to calculate moles of ethanol in 1 L of solution or nearly 1 L of water because the solution is dilute.
No. of moles of water in 1 L of water `=1000/8`
= 55.55 mole
`therefore (n(C_(2)H_(5)OH))/(n(C_(2)H_(5)OH) + 55.55) = 0.040`
`n(C_(2)H_(5)OH) = 0.04[n(C_(2)H_(5)OH)] + 2.222`
or `0.96 n (C_(2)H_(5)OH) = 2.222`
`therefore n=(2.222)/0.96 = 2.31`
`therefore 231` moles of ethanol are present in 1 L of solution and hence, molarity of the solution is 2.31 M.