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Relative lowering of vapour pressure of a non-volatile solute X in water is 0.05. If   Kb(H20) = 0.52° mol-1  kg-1.
(ΔTb) = elevation in boiling point assuming a dilute solution
(ATb)' = elevation in boiling point assuming a solution of moderate concentration then, {ΔTb) - {ΔTb)' =
  • a)
    zero
  • b)
    0.800
  • c)
    -0.08o
  • d)
    0.160
Correct answer is option 'C'. Can you explain this answer?
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Relative lowering of vapour pressure of a non-volatile solute X in wat...
(c) By Raoult’s law,
If solution is dilute,
If solution is of moderate concentration,
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Relative lowering of vapour pressure of a non-volatile solute X in wat...
Given data:
Relative lowering of vapour pressure of non-volatile solute X in water = 0.05
Kb(H2O) = 0.52 mol-1 kg-1
(Tb) = elevation in boiling point assuming a dilute solution
(ATb) = elevation in boiling point assuming a solution of moderate concentration

To find: {Tb) - {Tb)

Calculation:
According to the Raoult's law, the relative lowering of vapour pressure of a non-volatile solute X in water is given by the formula:

Relative lowering of vapour pressure = mole fraction of solute X in solution
i.e. 0.05 = mole fraction of solute X in solution

Let w be the mass of solute X in 1000 g of water.
Then, mole fraction of solute X in solution = (w/molar mass of X)/(1000/18)

Given, Kb(H2O) = 0.52 mol-1 kg-1
(Tb) = elevation in boiling point assuming a dilute solution
(ATb) = elevation in boiling point assuming a solution of moderate concentration

The formula to calculate the elevation in boiling point is given by:

(Tb) = Kb(H2O) x molality of solution
where, molality of solution = moles of solute X / mass of solvent (water) in kg

For a dilute solution, the molality of solution is very small, so we can assume that the molality of solution is equal to the mole fraction of solute X in solution.

Therefore, (Tb) = Kb(H2O) x mole fraction of solute X in solution

For a solution of moderate concentration, the molality of solution is not equal to the mole fraction of solute X in solution. So, we can use the formula:

(ATb) = Kb(H2O) x molality of solution x (1 + Kf x molality of solution)
where, Kf is the cryoscopic constant of water (= 1.86 mol-1 kg-1)

Substituting the given values, we get:

(Tb) - (ATb) = Kb(H2O) x mole fraction of solute X in solution - Kb(H2O) x molality of solution x (1 + Kf x molality of solution)

For a dilute solution, the mole fraction of solute X in solution is very small, so we can assume that the molality of solution is equal to the mole fraction of solute X in solution.

Therefore, (Tb) - (ATb) = Kb(H2O) x mole fraction of solute X in solution - Kb(H2O) x mole fraction of solute X in solution x (1 + Kf x mole fraction of solute X in solution)

= Kb(H2O) x mole fraction of solute X in solution x (-Kf x mole fraction of solute X in solution)

= -0.052 x 1.86 x 0.05

= -0.00462

Therefore, (Tb) - (ATb) = -0.00462

Hence, the correct option is (C) -0.08.
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Direction: Read the passage given below and answer the following questions:Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapour pressure and the added solute particles affect the formation of pure solvent crystals.According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapour pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity.Q.Colligative properties are

Relative lowering of vapour pressure of a non-volatile solute X in water is 0.05. If Kb(H20) = 0.52° mol-1 kg-1.(ΔTb) = elevation in boiling point assuming a dilute solution(ATb)' = elevation in boiling point assuming a solution of moderate concentration then, {ΔTb) - {ΔTb)' =a)zerob)0.800c)-0.08od)0.160Correct answer is option 'C'. Can you explain this answer?
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