alleryMole fraction of solute in 18% glucose (180 g/mol) by mass in water:
A0.02
B0.03
C0.1
D0.05
Answer & Solution
Correct answer: A. 0.02
**In 100 g solution:** 18 g glucose (= 0.1 mol), 82 g water (= 82/18 = 4.56 mol).
x_glucose = 0.1/(0.1 + 4.56) = 0.1/4.66 ≈ 0.0215 ≈ 0.02.
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