At `300K`, two solutions of glucose in water of concentration `0.01M` and `0.001M` are separated by semipermeable membrane. Pressure needs to be applied on which solution, to prevent osmosis? Calculate the megitude of this applied pressure?

Answer

`P = 0.2217` atm should be applied

Explanation

`{:(For 0.01M solution),(pi_(1)V_(1)=n_(1)S_(1)T_(1)),( :. pi_(1) =0.01 xx 0.0821 xx 300),(=0.2463 atm),(For 0.0001 M solution),(pi_(2)V_(2)=n_(2)ST_(2)),( :. pi_(2)=0.001 xx 0.0821 xx 300),(=0.02463 atm):}:|:{:(,),(n_(1)//V_(1)=0.01,),(T =300K,),(,),(n_(2)//V_(2)=0.001,),(T=300K,):}`
The movement of solvent particles occurs from dilute to concentrate solution i.e., `0.001M` to `0.01M` solution. Thus, pressure should be applied on concentrated solution i.e., on `0.01M` solution to prevent osmosis.
`= 0.2463 - 0.0246 = 0.2217 atm` pressure on `0.01M` solution/