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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 15:
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
. 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.An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Here,
Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar (Given)
Vapour pressure of pure water at normal boiling point (p10) = 1.013 bar
Mass of solute, (w2) = 2 g
Mass of solvent (water), (w1) = 100 - 2 = 98 g
Molar mass of solvent (water), (M1) = 18 g mol - 1
According to Raoult's law,
(p10 - p1) / p10 = (w2 x M1 ) / (M2 x w1 )
(1.013 - 1.004) / 1.013 = (2 x 18) / (M2 x 98 )
0.009 / 1.013 = (2 x 18) / (M2 x 98 )
M2 = (2 x 18 x 1.013) / (0.009 x 98)
M2 = 41.35 g mol - 1
Hence, the molar mass of the solute is 41.35 g mol - 1.
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 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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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 15: An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling p....
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Comments
there is a little mistake in formula right formula is (p1* - p1) / p1 = (w2 x M1 ) / (M2 x w1 ) by this formula the answer is 40.9 which is correct also mention in NCERT question 2.15
Very helpful
In the question,it's only given as a solvent.Why did u take it as water?
Thank you so much very informative
Thanks
Thx
Thanks
isme vapour presure water ka to diya hi na h
Thanks
Thanks