Question

Calculate the mole fraction of solute in the following aqueous solutions.

a) 22.3% C2H5OH, by mass

b) 0.694 m CO(NH2)2 (urea)

Answer 1

a) 22.3% C2H5OH, by mass

let us take 100 gm in which

22.3 gm ethanol + 77.7 gm water

Moles of ethanol = 22.3 /46.07 = 0.48415 Moles

Moles of water = 77.7 /18 = 4.316 Mole

Total mole = 4.8 Mole

Mole fraction of ethanol = 0.48415 / 4.8 = 0.1

Mole fraction of water = 4.316 / 4.8 = 0.8991

b) 0.694 m CO(NH2)2 (urea)

0.694 m x 60.06 = 41.68 gm of urea

let us take 100 gm in which

41.68 gm of urea + 58.32 gm of water

Moles of water = 58.32 / 18 = 3.24 Mole

Total moles 0.694 + 3.24 = 3.934

Mole fraction of urea = 0.694 / 3.934 = 0.1764

Mole fraction water = 3.24/ 3.934 = 0.8235