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 19:
A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
1) Molar mass of the solute
2) Vapour pressure of water at 298 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 solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:
1) Molar mass of the solute
2) Vapour pressure of water at 298 K.
Let, the molar mass of the solute be M g mol - 1
Now, the no. of moles of solvent (water),n1 = 90g / 18g mol-1
And, the no. of moles of solute,n2 = 30g / M mol-1 = 30 / M mol
p1 = 2.8 kPa
Applying the relation:
(p10 - p1) / p10 = n2 / (n1 + n2)
⇒ (p10 - 2.8) / p10 = (30/M) / {5 + (30/M)}
⇒ 1 - (2.8/p10) = (30/M) / {(5M+30)/M}
⇒ 1 - (2.8/p10) = 30 / (5M + 30)
⇒ 2.8/p10 = 1 - 30 / (5M + 30)
⇒ 2.8/p10 = (5M + 30 - 30) / (5M + 30)
⇒ 2.8/p10 = 5M / (5M+30)
⇒ p10 / 2.8 = (5M+30) / 5M ----------------(1)
After the addition of 18 g of water:
n1 = (90+18g) / 18 = 6 mol
and the new vapour pressure is p1 = 2.9 kPa (Given)
Again, applying the relation:
(p10 - p1) / p10 = n2 / (n1 + n2)
⇒ (p10 - 2.9) / p10 = (30/M) / {6 + (30/M)}
⇒ 1 - (2.9/p10) = (30/M) / {(6M+30)/M}
⇒ 1 - (2.9/p10) = 30 / (6M + 30)
⇒ 2.9/p10 = 1 - 30 / (6M + 30)
⇒ 2.9/p10 = (6M + 30 - 30) / (6M + 30)
⇒ 2.9/p10 = 6M / (6M+30)
⇒ p10 / 2.9 = (6M+30) / 6M ----------------(2)
Dividing equation (1) by (2),we get:
2.9 / 2.8 = {(5M+30) / 5M} / {(6M+30) / 6M}
⇒ 2.9 x (6M+30 / 6) = (5M+30 / 5) x 2.8
⇒ 2.9 x (6M +30) x 5 = (5M+30) x 2.8 x 6
⇒ 87M + 435 = 84M + 504
⇒ 3M = 69
⇒ M = 23u
Therefore, the molar mass of the solute is 23 g mol - 1.
(ii) Putting the value of 'M' in equation (i), we get:
⇒ p10 / 2.8 = (5M+30) / 5M
⇒ p10 / 2.8 = (5x23+30) / 5x23
⇒ p10 = (145 x 2.8) / 115
⇒ p10 = 3.53
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
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 19: A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of ....
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Comments
Thank you so much..it really helped me a lot
Why are you adding 1 in pð-p/pð ??
Nice
Really good