Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
It is given that:
Volume of water, V= 450 mL = 0.45 L
Temperature, T = (37 + 273)K = 310 K
Number of moles of the polymer,
n = 1 / 185000 mol
We know that:
Osmotic pressure,
= 1/185000 mol X 1/0.45L X 8.314 X 103 Pa L K-1mol-1 X 310K
= 30.98 Pa = 31 Pa (approximately)
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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 1 , Question 12: Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of poly....
Comments
After solving the numerical value the ans given is not coming
How is the value 10^3 came with the value of R
How is the value 10^3 came with the value of R
R=8.314JK^(-1)=8.314Ã10^3 PaL-1K-1
good,ok one of the best
How that 10 ^3 came with 8.314