Practice question

Given: Calculate the boiling point elevation for a solution of 17.1 g of sucrose ( C₁₂H₂₂O₁₁ ) in 200 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 the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.05/0.2 = 0.25 mol/kg . Δ T_b = 0.52 × 0.25 = 0.13 K . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.