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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
∴ Molality of the solution (m) = `(0.0146 "mol")/(0.095 kg)`
= 0.1537 mol kg−1
Applying the relation,
ΔTf = Kf × m
⇒ Kf = `(Delta T_f)/m`
= `(2.15 K)/(0.1537 "mol" kg^(-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
∴ 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)
Freezing point of the 5% glucose solution is = 273.15 − 4.09 K
= 269.06 K
Solution 2
For a cane sugar solution,
M' = `(1000 xx K_f xx w)/(W xx Delta T_f)`
⇒ 342 = `(1000 xx K_f xx 5)/(95 xx (273.15 - 271))`
342 = `(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)`
180 = `(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`
⇒ ΔTf = `(342 xx 2.15)/180`
= 4.085 K
Freezing point of glucose solution = 273.15 − 4.085
= 269.07 K
∴ The freezing point of a glucose solution is 269.07 K.
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| 4 | 0.6 | -3.8 | -3.8 |
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| 6 | 1.0 | -6.4 | -6.3 |
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