Practice question

Given: Calculate the boiling point elevation of a solution containing 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 450 g of water. ( K_b = 0.52 K kg/mol, Molar mass of sucrose = 342 g/mol ) These values define the system as per NCERT data. Formula: Moles of sucrose = 17.1/342 = 0.05 mol. This is standard NCERT relation. Substitution & Calculation: Molality = 0.05/0.45 = 0.111 mol/kg . Δ T_b = 0.52 × 0.111 = 0.058 K . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.