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: \[ \pi = i C R T \]
Step 1: Moles of solute. \[ \frac{0.088}{174} = 5.06 \times 10^{-4} \, mol \]
Step 2: Molarity. Volume = 2 L \[ C = 2.53 \times 10^{-4} \, M \]
Step 3: Van’t Hoff factor. \[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \Rightarrow i = 3 \]
Step 4: Substitute values. \[ \pi = 3 \times 2.53 \times 10^{-4} \times 0.082 \times 300 \] \[ \pi = 0.0187 \, atm \]
Final Answer: \[ \boxed{0.019 \, atm} \]