At 300 K, two solutions of glucose in water of concentration 0.01 M and 0.001 M are separated by semipemeable membrane. Pressure needs to be applied on solution, to prevent osmosis Calculate the magnitude of this applied pressure.
A
P=0.111atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
P=0.222atm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
P=0.264atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
P=0.528atm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is BP=0.222atm The osmotic pressure, Π=CRT=(0.01−0.001)×0.08206×300=0.222 atm Here, Π is the osmotic pressure, C is the difference in the concentration of two solutions, R is the ideal gas constant and T is the absolute temperature.