In a 66.0-g aqueous solution of methanol, CH 4 O, CH4O, the mole fraction of methanol is 0.300. 0.300. What is the mass of each component?

The solution is composed by 37.5 g of water and 28.5 g of methanol.

Explanation:

Total mass = 66 g

Mole fraction methanol: 0.3

Sum of mole fraction = 1

Therefore, mole fraction of water = 0.7

Let's find out the mass of each component by this two equations:

Methanol mass + Water mass = 66 g

Water mass = 66g - Methanol mass

Methanol mass = 66g - water mass

water mass / 18 g/mol = moles of water

methanol mass / 32 g/mol = moles of methanol

moles of water / total moles = 0.7

moles of methanol / total moles = 0.3

water mass / 18 g/mol / (water mass / 18 g/mol) + (methanol mass / 32 g/mol) = 0.7; let's replace methanol mass, as (66 - water mass)

water mass / 18 g/mol / (water mass / 18 g/mol) + (66 - water mass / 32 g/mol) = 0.7 → The unknown is water mass (X)

X / 18 / ((X / 18) + ((66-X) / 32)) = 0.7

(X / 18) + ((66-X) / 32) = (16X + 594 - 9X) / 288

X / 18 / (16X + 594 - 9X) / 288 = 0.7

X/18 = 0.7. ((7X + 594) / 288)

X / 18 = 7/2880 (7X + 594)

X = 7/2880 (7X + 594) 18

X = 7/160 (7X + 594)

X = 49/160X + 2079/80

X - 49/160X = 2079/80

111/160X = 2079/80

X = 2079/80. 160/111 = 37.5 g → mass of water

Therefore, mass of methanol → 66 g - 37.4 g = 28.5 g