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Mole fraction of solute in a solution containing 18 g of glucose (M = 180) in 90 g of water is
A0.05
B0.100
C0.500
D0.020
Answer & Solution
Correct answer: D. 0.020
1. Moles of glucose n_g = 18/180 = 0.1 mol.
2. Moles of water n_w = 90/18 = 5 mol.
3. x_glucose = n_g / (n_g + n_w) = 0.1 / 5.1 ≈ 0.020.
_Source: Maharashtra Balbharati Std XII Chemistry, Ch 2 "Solutions" §2.3_
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