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प्रश्न
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.
उत्तर
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 = `5/342` mol
= 0.0146 mol
Therefore, molality of the solution, m = `(0.0146 "mol")/(0.095 "kg")`
= 0.1537 mol kg−1
Applying the relation,
ΔTf = Kf × m
`=> "K"_"f" = (triangle"T"_"f")/"m"`
`=> "K"_"f" = (2.15 "K")/(0.1537 "mol kg"^(-1))`
Kf = 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 the 5% glucose solution is (273.15 − 4.09) K = 269.06 K.
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