Answer:
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
or
ΔTf= Kfm
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.
What is meant by positive deviations from Raoult's law? Give an example. What is the sign of mixH for positive deviation?
OR
Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult's law? Give an example.
3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated).
(Given: Molar mass of benzoic acid = 122 g mol-1, Kf for benzene = 4.9 K kg mol-1
Calculate the mass of compound (molar mass = 256 g mol-1) to be dissolved in 75 of benzene to lower its freezing point by 0.48 K (Kf= 5.12 kg mol-1 )
18 g of glucose, C6H12O6 (Molar Mass = 180 g mol-1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil?
(Kb for water = 0.52 K kg mol-1, boiling point of pure water = 373.15 K)
Determine the osmotic pressure of a solution prepared by dissolving 2.5 x 10-2 g of K2SO4 in 2L of water at 250C, assuming that it is completely dissociated.
(R = 0.0821 L atm K-1 mol-1, Molar mass of K2SO4 = 174 g mol-1)