Table of Contents
Osmosis:
The phenomenon of osmosis was studied for the first time by Abbe Nollet in 1748. Let us consider an aqueous solution of sugar placed in an inverted thistle funnel having a semi-permeable membrane such as animal bladder or parchment paper, attached to its bottom. The thistle funnel is lowered into a beaker containing water. The membrane allows only the molecules of the solvent and not of the solute to pass through it. In other words, the membrane is permeable only to solvent molecules from pure solvent into the solution. As a result, water passes into the thistle funnel and the level of solution in the thistle funnel rises gradually. This process is called Osmosis. Thus,
| The phenomenon of the flow of solvent through a semipermeable membrane from pure solvent to the solution is called osmosis. |
Osmosis can also take place between the solutions of different concentrations. In such cases, the solvent molecules move from the solution of low solute concentration to that of higher solute concentration.
Semi-permeable membrane– A semi-permeable membrane is one that allows only the solvent and not the solute to pass through it. Nature has provided many such membranes both in plants and animals for specific functions. Example- parchment, cellophane membranes etc. They are not particularly useful in the laboratory due to their imperfect nature. Consequently, they can be even artificially prepared. One such membrane is the film of gelatinous precipitates of cupric ferrocyanide, Cu2[Fe(CN)6].
Preparation of Semi-permeable Membrane:
The best semipermeable membrane is cupric ferrocyanide Cu2[Fe(CN)6]. This is prepared by the reaction between CuSO4 and K2[Fe(CN)6].
| 2Cu+2 + [Fe(CN)6]-4 ————-> Cu2[Fe(CN)6] |
Osmotic Pressure:
To understand the concept of osmotic pressure, consider an apparatus shown in the figure. It consists of a vessel divided into two compartments by a semi-permeable membrane. These two compartments are fitted with water-tight frictionless pistons. Let us take solution in one compartment and pure solvent in the other compartment. Due to osmosis, there will be a flow of solvent into the solution compartment through the semi-permeable membrane. As a result, the piston on the solution side will tend to move outwards. To stop this movement of the piston outwards, we have to apply pressure on the solution side. This pressure just sufficient to stop osmosis will be equal to the osmotic pressure. The osmotic pressure may be defined as,
| The excess pressure which must be applied to a solution to prevent the passage of solvent into it through a semi-permeable membrane. |
Thus, osmotic pressure is the pressure applied to the solution to prevent osmosis. It is generally denoted by π.
Osmotic pressure- a Colligative Property:
Van’t Hoff (1887) concluded that a dilute or ideal solution behaves like an ideal gas and the different gas laws are applicable to the dilute solutions as well.
Van’t Hoff observed that for dilute solutions, the osmotic pressure (π) is given as-
| π = cRT |
where c is the molar concentration of the solution (molarity), T is the temperature and R is the gas constant.
For a solution, at a given temperature, both R and T are constant.
| ∴ π ∝ c |
Since osmotic pressure depends upon the molar concentration of solution it is, therefore, a colligative property.
Derivation of Osmotic Pressure (π):
| According to Boyle’s Vant Hoff law, At constant temperature, π ∝ C ……….(i) If, C = 1/V, then equation (i) becomes, π ∝ 1/V and πV = constant According to Charle’s Vant Hoff Law, When concentration is constant, π ∝ T or, π/T = constant ……….(ii) By taking both the laws, we have, π ∝ CT or, π = CRT ……….(iii) Where, C = n/V Put the value of C in equation (iii) π = (n/V) RT or, πV = nRT |
Osmotic pressure (π) is a colligative property because the osmotic pressure of a dilute solution depends upon the number of moles of solute present in the solution and not upon the composition.
Determination of Molar Mass of Solute from Osmotic Pressure:
| According to Van’t Hoff Equation, π = cRT But c= n/V Where n is the number of moles of solute dissolved in V litres of the solution. ∴ π = (n/V)RT The number of moles of solute n may be given as wB/MB. Here wB is the weight of the solute and MB is its molar mass. Substituting the value of n in the above expression, π = wBRT/VMB or MB = wBRT/Vπ Thus the molar mass of the solute, MB, can be calculated. |
