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

Part (i): Calculation of osmotic pressure

Concept: Osmotic pressure is a colligative property that depends on the number of solute particles present in the solution. For an electrolyte: \[ \boxed{\pi=iCRT} \] where, \[ \pi=\text{osmotic pressure} \] \[ i=\text{Van't Hoff factor} \] \[ C=\text{molar concentration} \] \[ R=\text{gas constant} \] \[ T=\text{temperature in Kelvin} \]

Step 1: Calculate Van't Hoff factor Potassium sulphate dissociates completely: \[ K_2SO_4\rightarrow2K^++SO_4^{2-} \] Number of ions produced: \[ 2+1=3 \] Therefore: \[ \boxed{i=3} \]

Step 2: Calculate moles of $K_2SO_4$ \[ \text{Moles}=\frac{\text{Mass}}{\text{Molar mass}} \] \[ =\frac{0.025}{174} \] \[ =1.437\times10^{-4}mol \]

Step 3: Calculate molarity Volume of solution: \[ V=2L \] Therefore: \[ C=\frac{\text{moles of solute}}{\text{volume of solution}} \] \[ C=\frac{1.437\times10^{-4}}{2} \] \[ C=7.18\times10^{-5}M \]

Step 4: Calculate osmotic pressure Temperature: \[ T=27+273=300K \] Using: \[ \pi=iCRT \] Substituting values: \[ \pi=3\times7.18\times10^{-5}\times0.082\times300 \] \[ \pi=5.29\times10^{-3}atm \] Hence: \[ \boxed{\pi=5.29\times10^{-3}atm} \]

Part (ii): Azeotrope formed by acetone and chloroform

Concept: An azeotrope is a liquid mixture that boils at a constant temperature and has the same composition in liquid and vapour phases. Azeotropes are of two types:

• Minimum boiling azeotrope: Shows positive deviation from Raoult's law.

• Maximum boiling azeotrope: Shows negative deviation from Raoult's law.
In acetone-chloroform mixture, hydrogen bonding occurs between hydrogen atom of chloroform and oxygen atom of acetone: \[ CHCl_3\cdots O=C(CH_3)_2 \] This strong interaction reduces the escaping tendency of molecules. Therefore: Vapour pressure decreases and: Boiling point increases Hence, the mixture shows negative deviation from Raoult's law and forms a: \[ \boxed{\text{Maximum boiling azeotrope}} \]

Note: Acetone-chloroform actually forms a: \[ \boxed{\text{Maximum boiling azeotrope}} \] because of strong intermolecular hydrogen bonding and negative deviation from Raoult's law.

Final Answer:

(i) \[ \boxed{\pi=5.29\times10^{-3}atm} \]

(ii) Acetone and chloroform form a: \[ \boxed{\text{Maximum boiling azeotrope}} \] due to strong hydrogen bonding between acetone and chloroform molecules, causing negative deviation from Raoult's law.