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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 22:
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
. 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.At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Here we have given
π1= 4.98
π2 = 1.52
C1 = 36/180
C2 = ? (we have to find)
Now according to van’t hoff equation
Π = CRT
Putting the values in above equation,we get
4.98 = 36/180RT ------------------------1
1.52 = c2RT ------------------------2
Now dividing equation 2 by 1 ,we get
(c2 x 180) / 36 = 1.52 / 4.98
or
c2 = 0.061
Therefore concentration of 2nd solution is 0.061 M
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 22: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If....
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Comments
Nyc
C1=mole/volume=weight/molecular weight Ãâ 1/volume = 36/180 Ãâ 1/1
what is the meaning of writing C1= 36/180 but its real value is 36g
How can write 36/180
Sir its 0.061
How write 36/180
Sorry... The answer of this question is... 0.06 not 0.006