SolutionsNEET Chemistry · Class 12 · NCERT Chapter 1

Try this

• Try 10 g of glucose in 90 g of water (density = 1.04 g/mL). Molality uses only the 90 g of water; molarity uses the total solution volume.
• Change the solvent to benzene (M = 78). Notice the mole fraction changes even though the masses stay the same.
• Set mass of solute to 58.5 g, molar mass to 58.5 (NaCl), mass of solvent to 1000 g, density to 1.035. You get molality = 1 m and molarity just under 1 M.

Try this

• For ideal benzene + toluene: set χA = 0.5. Total pressure is the exact midpoint between P°A and P°B. This is the straight-line Raoult behaviour.
• Switch to positive deviation and move χA to 0.5. Notice the total pressure rises above the ideal prediction. The peak is the minimum boiling azeotrope composition.
• Compare mole fraction of A in vapour vs in liquid. Vapour is always richer in the more volatile component (higher P°). This is the basis of fractional distillation.

Try this

• Set i = 1 (urea/glucose), then change to i = 2 (NaCl). The ΔTf doubles. This shows why electrolytes lower the freezing point more than non-electrolytes at the same molality.
• Switch solvent to camphor (Kf = 40). A 0.01 m solution gives ΔTf = 0.4 K, easily measurable. This is why camphor is used in the Rast method for molar mass determination.
• Compare BaCl2 (i = 3) and Na3PO4 (i = 4) at the same molality. Na3PO4 gives larger colligative effects because it releases more ions.