A solution 0.1M $N{{a}_{2}}S{{O}_{4}}$ is dissolved to the extent of 95%. What would be its osmotic pressure at ${{27}^{o}}C$ ? (R = $0.0821Latm{{K}^{-1}}mo{{l}^{-1}}$ )

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Hint: When a non-volatile solute is added to a volatile solvent, the vapor pressure of the solution decreases. More properties are related to this decrease in vapor pressure. Those are relative lowering of vapour pressure, depression of freezing point, the elevation of the boiling point of the solvent, and osmotic pressure of a solution.

Complete step by step solution:
The process of flow of solvent molecules from pure solvent to the solution through a semipermeable membrane that passes solvent molecules only is known as the Osmosis process. The flow will continue in the process until the equilibrium is reached. The pressure required to stop the flow of solvent molecules through the semipermeable membrane is osmotic pressure.
Osmotic pressure depends on the number of solute particles because this is a colligative property. So, experimentally found that osmotic pressure is proportional to the molarity of the solution at a given temperature.
Thus, $\Pi =CRT$ ------ (1)
Where $\Pi $ = osmotic pressure
R = gas constant
C = molarity of solution
T = temperature
If i = van’t Hoff factor, then osmotic pressure $\Pi =iCRT$ -- (2)
Given, concentration of $N{{a}_{2}}S{{O}_{4}}$ (C) = 0.1M
Temperature (T) = ${{27}^{0}}C$ = 27+273K = 300K
Degree of dissociation of $N{{a}_{2}}S{{O}_{4}}$ , $\alpha $ = 0.95
Van’t Hoff factor, i = $1+\alpha (n-1)$
For$N{{a}_{2}}S{{O}_{4}}$, n =3
Then, i =1+0.95(3-1) = 2.9
Substitute the values i, C, and T in equation (2)
$\Pi =2.9\times 0.1\times 0.0821\times 300$ = 7.14 atm

Hence, the osmotic pressure ${{27}^{o}}C$ is 7.14atm.

Note: Another method of determining molar masses of solutes is measuring osmotic pressure. This method is applied to determine the molar masses of proteins, polymers, and macromolecules. Determination of molar masses of biomolecules which are not stable at high temperature is possible by this technique.