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Asked by Brunda Last Modified
Answer Here, 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
= 0.0146 mol
Therefore, molality of the solution, 
= 0.1537 mol kg−1
Applying the relation,
ΔTf = Kf × m
= 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 
= 0.0278 mol
Therefore, molality of the solution, 
= 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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