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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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....
Comments
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C1=mole/volume=weight/molecular weight à 1/volume = 36/180 à 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