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Academic session 2026-27, Chapter 1 in the current curriculum.
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter Solutions. This page offers a step-by-step solution to the specific question from Exercise 2, Question 20:
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.
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.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.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 20: A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing p....
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Comments
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Kf for water and sugar cane will be same for kf of water and glucose
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