A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
In case of cane sugar:
ΔTf = (273.15 - 271) K = 2.15 K
Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g mol - 1
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 - 5)g = 95 g of water.
Now, number of moles of cane sugar =5/342 mol
= 0.0146 mol
Therefore, molality of the solution,m =0.0146mol / 0.095kg
= 0.1537 kg mol - 1
Now applying the relation,
ΔTf = Kf × m
⇒ Kf = ΔTf / m
⇒ 2.15K / 0.1537 kg mol-1
= 13.99 K kg mol-1
Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol - 1
5% glucose in water means 5 g of glucose is present in (100 - 5) g = 95 g of water.
∴ Number of moles of glucose = 5/180 mol
= 0.0278 mol
Therefore, molality of the solution,m =0.0278 mol / 0.095 kg
= 0.2926 mol kg - 1
Applying the relation,
ΔTf = Kf × m
= 13.99 K kg mol - 1 × 0.2926 mol kg - 1
= 4.09 K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 - 4.09) K= 269.06 K.
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8 Comment(s) on this Question
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Nice answer
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Kf for water and sugar cane will be same for kf of water and glucose
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