# Osmosis and Osmotic Pressure

The phenomenon of the passage of pure solvent from a region of lower concentration (of the solution) to a region of its higher concentration through a semi-permeable membrane is called osmosis.

The origin of osmosis can be explained as that solvent molecules can pass readily through pores in the membrane into the solution. However, the solute molecules block the return of some solvent molecules into the pure solvent.
As a result flow from the pure solvent into the solution is initially faster than return flow. The return flow increases as growing pressure forces more solvent molecules back through the membrane more quickly.

The pressure (of the column of the solution developed on more concentrated side) at which both rates become equal is known as osmotic pressure.It is driving force of osmosis.

It is the difference in the pressure between the solution and the solvent system or it is the excess pressure which must be applied to a solution in order to prevent flow of solvent into the solution through the semi-permeable membrane.

Once osmosis is complete the pressure exerted by the solution and the solvent on the semi-permeable membrane is same.

Van’t Hoff equation for dilute solutions is ( parallel to ideal gas equation)

πV = nRT

where
π = Osmotic pressure
V = volume of solution
n = no. of moles of solute that is dissolved
R = Gas constant
T = Absolute temperature

Isotonic Solutions:
A pair of solutions having same osomotic pressure is called isotonic solutions

Illustration : A solution containing 8.6g per dm3 of urea (molar mass = 60g mol-1) was found to be isotonic with a 5 percent solution of an organic non-volatile solute. Calculate the molar mass of the latter.

Solution: According to the van’t Hoff theory, isotonic solutions have the same osmotic pressure at the same temperature and the same molar concentration.

Molar concentration of urea solution = 8.6 gdm-3/60 gmol-1

Let M2 be the molar mass of the unknown solute

∴ Molar concentration of the unknown solution = 50 g dm-3/M2

Since both the solutions are isotonic, hence, by definition, their molar concentrations are equal. Thus,

$\large \frac{50 g \; dm^{-3}}{M_2} = \frac{8.6 g \;dm^{-3}}{60 g \;mol^{-1}}$

M2 = 348.8 g mol–1