A solution of glucose (C6H12O6) was prepared by dissolving a certain amount in 100 g of water. The depression in freezing point is 0.0410 K. If molal depression constant for water is 1.86 K kg mol-1, calculate the weight of glucose dissolved. (C = 12, H=1,0 = 16) (Ans. 0.40 g)
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0.4 g of glucose
Explanation:
Given: Kf = 1.86 kg mol-1, molar mass of glucose (M₂) = 180 g/mol
W₂ = ?, ΔTf = 0.0410 K, W₁ = 100 g
We know that, Freezing point
ΔTf = Kf × W₂ × 1000/M₂ × W₁
W₂ = ΔTf × M₂ × W₁/ Kf × 1000
= 0.0410 × 180 × 100/1.86 × 1000
= 738/1860
= 0.396 ≅ 0.4 g of glucose
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