A solution is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9.
What is the molality `(m)` of the solution.
b. Water is added to the above solution such that the mole fraction of water in the solution becomes 0.9. What is the molality `(m)` of the solution?

Explanation

a. Molar mass of `H_(2) O = 18 mol^(-1)`
Molar mass of ethanol `(C_(2) H_(5) OH)`
`= 2 xx 12 + 5 + 16 + 1`
`= 46 g mol^(-1)`
Mole fraction of ethanol is 0.9 and is greater than the mole fraction of water (i.e., 1 - 0.9 = 0.1), so ethanol is solvent and water is solute. Thus `X_(1) = 0.9 ` and `X_(2) = 0.1 (Mw_(1) = 46` and `Mw_(2) = 18)`.
`:. m = (X_(2) xx 1000)/(X_(1) xx Mw_(1)) = (("Mole fraction of solute" xx 1000)/("Mole fraction of solvent" xx "Molar mass of solvent"))`
`(0.1 xx 1000)/(0.9 xx 46) = 2.415 m`
b. Now mole fraction of water is `0.9` and is greater than the mole fraction of ethanol (i.e, 1 - 0.9 = 0.1), so water is solvent and ethanol is solute. Thus, `X_(1) = 0.9, X_(2) = 0.1 (Mw_(1) = 18` and `Mw_(2) = 46)`.
`:. m = (X_(2) xx 100)/(X_(1) xx Mw_(1)) = (0.1 xx 1000)/(0.9 xx 18) = 6.17 m`