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Question
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.
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Solution 1
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 = (Delta 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.
Solution 2
For cane sugar solution,
`M' = (1000 xx K_f xx w)/(W xx Delta T_f)`
or `342 = (1000 xx K_f xx 5)/(95 xx (273.15 - 271))`
= `(5 xx 1000 xx K_f)/(95 xx 2.15)` ...(i)
For glucose solution,
`180 = (1000 xx K_f xx 5)/(95 xx Delta T_f)`
= `(5 xx 1000 xx K_f)/(95 xx Delta T_f)` ...(ii)
Dividing eq. (i) by eq. (ii), we have
`342/180 = (Delta T_f)/2.15`
or `Delta T_f = (342 xx 2.15)/180`
= 4.085 K
Hence, freezing point of glucose solution = 273.15 − 4.085 = 269.07 K
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