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The mole fraction of ethylene glycol (C₂H₆O₂, M = 62) in a 20% by mass aqueous solution is approximately:
A0.068
B0.50
C0.20
D0.30
Answer & Solution
Correct answer: A. 0.068
In 100 g: 20 g glycol (→ 20/62 = 0.322 mol) + 80 g water (→ 80/18 = 4.444 mol). x_glycol = 0.322/(0.322+4.444) = **0.068**. x_water = 0.932.
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