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  • 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?

At 300 K, 36 g of glucose present in a l | Class 12 Chemistry Chapter Solutions, Solutions NCERT Solutions

Q22.

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

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Important Questions & Answers

What is the correct answer to: 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 -...

How do you solve 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? step by step?

Step-by-step explanation:
• Here we have given
•
• π1= 4
• 98
•

Why is this answer important for exams?

This question is important because it tests key concepts from the NCERT syllabus and is frequently asked in CBSE exams.

Which NCERT concept is used in this question?

This question is based on core NCERT concepts explained in the chapter and should be revised thoroughly before exams.

What common mistakes should be avoided in this question?

Students often lose marks by skipping steps, writing incomplete explanations, or misunderstanding keywords used in the question.

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Student Discussion

Nevil
Class · · , · Aug 24, 2019
Nyc
👍 👎 Reply
Uzair shaikh
Class · · , · Jun 09, 2019
C1=mole/volume=weight/molecular weight × 1/volume = 36/180 × 1/1
👍 👎 Reply
manisha
Class · · , · May 28, 2019
what is the meaning of writing C1= 36/180 but its real value is 36g
👍 👎 Reply
Khushi
Class · · , · May 15, 2019
How can write 36/180
👍 👎 Reply
Obd
Class · · , · May 14, 2019
Sir its 0.061
👍 👎 Reply
RAUSHAN KUMAR
Class · · , · Apr 10, 2019
How write 36/180
👍 👎 Reply
Chandan Sharma
Class · · , · Sep 17, 2018
Sorry... The answer of this question is... 0.06 not 0.006
👍 👎 Reply

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