Practice question

Given: A solution of 12 g of a non-volatile solute in 100 g of water boils at 100.52°C at 1 atm. 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: Δ T_b = 100.52 - 100 = 0.52 K. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.52/0.52 = 1 mol/kg . Moles = 1 × 0.1 = 0.1 mol . Molar mass = 12/0.1 = 120 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.