Practice question

Given: What is the freezing point depression of a solution containing 3.6 g of glucose ( C₆H₁₂O₆ ) in 100 g of water? ( K_f = 1.86 K kg/mol, Molar mass of glucose = 180 g/mol ) These values define the system as per NCERT data. Formula: Moles of glucose = 3.6/180 = 0.02 mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Molality = 0.02/0.1 = 0.2 mol/kg . Δ T_f = 1.86 × 0.2 = 0.372 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.