Question

A solution is prepared by dissolving 0.025 g of potassium sulphate in 2 L of water at 27$^\circ$C. Assuming potassium sulphate is completely dissociated, determine its osmotic pressure. Given: \[ R=0.082 \, L\,atm\,K^{-1}mol^{-1} \] \[ \text{Molar mass of }K_2SO_4=174\,g\,mol^{-1} \] (ii) What type of azeotrope will be formed by a solution of acetone and chloroform? Give reason.

Hint:

For electrolytes: \[ \boxed{\pi=iCRT} \] Always include the Van't Hoff factor. Azeotrope shortcut: \[ \boxed{\text{Positive deviation} \Rightarrow \text{Minimum boiling azeotrope}} \] \[ \boxed{\text{Negative deviation} \Rightarrow \text{Maximum boiling azeotrope}} \] Acetone + chloroform: \[ \boxed{\text{Strong H-bonding} \Rightarrow \text{Maximum boiling azeotrope}} \]

Answer 1

Step 1: Dissociation and concentration of $K_2SO_4$.
$K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$, so $i = 3$. Moles $= \frac{0.025}{174} = 1.437 \times 10^{-4}\,mol$. Molarity $C = \frac{1.437 \times 10^{-4}}{2\,L} = 7.18 \times 10^{-5}\,M$.
Step 2: Osmotic pressure calculation.
$T = 27 + 273 = 300\,K$. Using $\pi = iCRT$: \[ \pi = 3 \times 7.18 \times 10^{-5} \times 0.082 \times 300 = 5.29 \times 10^{-3}\,atm \]
Step 3: Interaction in acetone-chloroform mixture.
$CHCl_3$ and acetone form intermolecular hydrogen bonds ($CHCl_3 \cdots O=C(CH_3)_2$). This interaction is stronger than those in pure components, reducing the escaping tendency of molecules and lowering vapour pressure.
Step 4: Type of azeotrope formed.
Reduced vapour pressure means negative deviation from Raoult's law ($\Delta H_{mix} < 0$, $\Delta V_{mix} < 0$). Therefore acetone-chloroform forms a maximum boiling azeotrope.