A tube of uniform cross-sectional area `1 cm^(2)` is closed at one end with semi-permeable membrane. A solution of `5 g` glucose per `100 mL` is placed inside the tube and is dipped in pure water at `27^(@)C`. When equilibrium is established, calculate:
a. The osmotic pressure of solution.
b.The height developed in vertical column.
Assume the density of final glucose solution `1 g mL^(-1)`

Explanation

a. `piV=W/(Mw)(RT)`
Given that `W= 5 g`, `V=100/1000 L`
`T=300 K`,`Mw= 180` (Mw of glucose =180)
`pi=(5 xx 1000 xx 0.821 xx300)/(180 xx 100)=6.842` atm
b. `pi=hdg`
Given `d=1 g cm^(-3) =10^(3) kg m^(-3)`
`pi=6.842 = 6.842 xx 1.01 xx 10^(5) N m^(-2)`
`g=9.81 m s^(-2)`
`:. 6.841 xx 1.01 xx10^(5) =h xx 10^(3) xx 9.81`
`h=(6.841 xx 1.01 xx 10^(5))/(9.81 xx 10^(3)) =70.43 m`