Practice question

Given: A solution of 4.8 g of a non-volatile solute in 150 g of water has a boiling point elevation of 0.208 K. What is the molar mass of the solute? ( K_b = 0.52 K kg/mol ) These values define the system as per NCERT data. Formula: Molality = Δ T_b/K_b = 0.208/0.52 = 0.4 mol/kg. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Moles = 0.4 × 0.15 = 0.06 mol . Molar mass = 4.8/0.06 = 80 g/mol . 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.