(a) Define the following terms :
(i) Mole fraction.
(ii) Ideal solution.
(b) 15.0 g of an unknown molecular material was dissolved in 450 g of water. The resulting solution was found to freeze at -0.34 °C. What is the molar mass of this material? (K_ƒ for water = 1.86 K kg mol^-1 )
OR
(a) Explain the following:
(i) Henry’s law about dissolution of a gas in a liquid.
(ii) Boiling point elevation constant for a solvent.
(b) A solution of glycerol (C₃H₈O₃) in water was prepared by dissolving some glycerol in 500 g of water. This solution has a boiling point of 100.42°C. What mass of glycerol was dissolved to make this solution? (KA for water = 0.512 K kg mol-1).

(a)

(i) Mole fraction is a way of expressing the concentration of a solution. It represents the number of molecules of a particular component in a mixture divided by the total number of moles in the given mixture.

(ii) An ideal solution is one in which the molecules attract one another with equal force irrespective of their nature. Thus, a solution composed of two components A and B will be an ideal one if the forces between A and A, B and B should be the same.

An ideal solution possesses the following characteristics:

(i)Volume change of mixing should be zero.

(ii)Heat change on mixing should be zero.

(iii)There should be no chemical reaction between solvent and solute.

(iv)Solute molecules should not dissociate in the ideal solution.

(v)Solute molecules should not associate in the ideal solution.

(vi)Ideal solutions must obey Raoult’s law at all concentrations.

Some of the binary mixtures which show the properties of ideal solutions:-

(a) Benzene and toluene.

(b) n-Hexane and n-heptane.

(b) We know depression in freezing point ΔT_f =

Where K_f is the molal freezing point depression constant of solvent = 1.86 K kg mol^-1

W₂ = weight of unknown molecular material = 15gm.

W₁ = weight of water given = 450gm.

ΔT_f = change in freezing point = 0.34°C

M₂ = molecular mass of the material.

∴ ΔT_f =

So, M₂ = = = 182.35gm mol^-1.

So, molar mass of the material is 182.35gm.

OR

(a)

(i) Henry’s law states that the partial pressure of the gas in vapour phase is directly proportional to the mole fraction of the gas dissolved in the liquid (in the solution).

p = K_Hx

K_H is Henry’s law constant.

(ii) Boiling point elevation constant is numerically equal to elevation in boiling point of 1 molal solution (moles of solute is dissolved in 1kg of solvent). It is denoted by K_b.

K_b =

Where K_b is the boiling point elevation constant of solvent.

W₂ = weight of solute.

W₁ = weight of solvent.

ΔT_f = change in boiling point.

M₂ = molecular mass of the solute.

(b) We know,

We know elevation in boiling point ΔT_b =

Where K_b is the boiling point elevation constant of solvent.

W₂ = weight of solute.

W₁ = weight of solvent.

ΔT_f = change in boiling point.

M₂ = molecular mass of the solute.

Given,

ΔT_b = 0.42°C.

K_b = 0.512 K kg mol^-1.

W₁ = 500gm.

M₂ = Molecular mass of glycerol = 92gm mol^-1.

So, W₂ = = = 37.73gm.

37.73 gm of glycerol was required to make this solution.