Question

Boiling point of water at \(750\,\mathrm{mmHg}\) is \(99.63^\circ\mathrm{C}\). The amount of sucrose to be added to \(500\,\mathrm{g}\) water so that it boils at \(100^\circ\mathrm{C}\) is_ _ _ _ g. (Molar elevation constant \(K_b = 0.5\,\mathrm{K\,kg\,mol^{-1}}\))

Hint:

Boiling point elevation depends only on number of particles (colligative property), not on type of solute.

Answer 1

Correct Answer: 127

Concept: Elevation in boiling point is a colligative property, which depends only on the number of solute particles, not their nature. When a non-volatile solute (like sucrose) is added to a solvent (water), it lowers the vapour pressure. Hence, a higher temperature is required for the solution to boil. \[ \Delta T_b = K_b \cdot m \] where:
•\(\Delta T_b\) = elevation in boiling point
•\(K_b\) = molal elevation constant
•\(m\) = molality \(= \frac{\text{moles of solute}}{\text{kg of solvent}}\)

Step 1: Required elevation in boiling point: \[ \Delta T_b = 100 - 99.63 = 0.37\,K \]

Step 2: Calculate molality: \[ m = \frac{\Delta T_b}{K_b} = \frac{0.37}{0.5} = 0.74 \,\text{mol/kg} \]

Step 3: Mass of solvent: \[ 500\,\text{g} = 0.5\,\text{kg} \] \[ n = m \times \text{mass} = 0.74 \times 0.5 = 0.37\,\text{mol} \]

Step 4: Molar mass of sucrose: \[ M = 342\,\text{g/mol} \]

Step 5: Mass required: \[ \text{mass} = n \times M = 0.37 \times 342 = 126.54 \approx 127\,g \] Final: \[{127\,g} \]