Calculate The Mole Fraction Of Ethylene Glycol
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20% of C2H6O2 by mass is present.
That means solution has 20 g of ethylene glycol and 80 g of water.
Now,
Molar mass of C2H6O2 = 12 x 2 + 1 x 6 + 2 x 16 = 62 g mol-1 .
Moles of C2H6O2 = 20 / 62 = 0.322 moles
Moles of water = 80 / 18 = 4.444 moles
Mole fraction of ethylene glycol = 0.322 / 0.322 + 4.444 = 0.068
Mole fraction of water = 1 - 0.068 = 0.932
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