Question

A solution is prepared by dissolving 0.088 g of potassium sulphate in 2 L of water at 27\(^\circ\)C. Assuming complete dissociation, determine its osmotic pressure. (Given: \(R = 0.082 \, L\,atm\,K^{-1}\,mol^{-1}\), Molar mass of K\(_2\)SO\(_4\) = 174 g mol\(^{-1}\))

Hint:

For osmotic pressure always convert temperature to Kelvin.

Answer 1

Concept:
Osmotic pressure is given by the formula: \[ \pi = i C R T \] where:

• \(\pi\) = osmotic pressure
• \(i\) = Van’t Hoff factor
• \(C\) = molarity of the solution
• \(R\) = ideal gas constant
• \(T\) = temperature in Kelvin

Step 1: Calculate moles of solute.
The number of moles of the solute is calculated by dividing the given mass of solute by its molar mass: \[ \frac{0.088}{174} = 5.06 \times 10^{-4} \, \text{mol} \]
Step 2: Calculate molarity.
The molarity is calculated by dividing the moles of solute by the volume of solution (in liters): \[ C = \frac{5.06 \times 10^{-4}}{2} = 2.53 \times 10^{-4} \, \text{M} \]
Step 3: Van’t Hoff factor.
For \(K_2SO_4\), the dissociation is: \[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \] Thus, the Van’t Hoff factor (\(i\)) is 3, as it dissociates into 3 ions.
Step 4: Substitute the values into the osmotic pressure equation.
Now, substitute all the known values into the osmotic pressure equation: \[ \pi = 3 \times 2.53 \times 10^{-4} \times 0.082 \times 300 \] \[ \pi = 0.0187 \, \text{atm} \]
Final Answer:
\[ \boxed{0.019 \, \text{atm}} \]