the freezing point of a substance may be defined as the temperature at which the vapour pressure of the substance in its liquid phase is equal to its vapour pressure in the solid phase.
According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and now it would become equal to that of solid solvent at lower temperature. Thus, the freezing point of the solvent decreases.
Let be the freezing point of pure solvent and Tf be its freezing point when non-volatile solute is dissolved in it. The decrease in freezing point.
Similar to elevation of boiling point, depression of freezing point (ΔTf) for dilute solution (ideal solution) is directly proportional to molality,
m of the solution. Thus,
ΔTf ∝ m
The proportionality constant, Kf, which depends on the nature of the solvent is known as Freezing Point Depression Constant or Molal Depression Constant or Cryoscopic Constant. The unit of Kf is K kg mol-1
If w2 gram of the solute having molar mass as M2, present in w1 gram of solvent, produces the depression in freezing point ΔTf of the
solvent then molality of the solute is given by the equation:
Thus for determining the molar mass of the solute we should know the quantities w1, w2, ΔTf, along with the molal freezing point depression constant.
Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1.
Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Let the total mass of the solution be 100g and mass of benzene be 30 g
therefore mass of tetrachloride= (100-30)g = 70g
Molar mass of benzene,
Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent and is expressed as:
Molarity (M) is defined as number of moles of solute dissolved in one litre (or one cubic decimetre) of solution.
(a) Mol. mass of
Volume of solution = 4.3 L
(b) Number of moles present in 1000 ml of 0.5M H2SO4= 0.5 mol
therefore number of moles present in 30ml of 0.5M H2SO4=mol =0.015mol
therefore molarity =0.015/0.5L
thus molarity is 0.03M