(a) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.

Given :

(Molar mass of sucrose = 342 g mol^–1)

(Molar mass of glucose = 180 g mol^–1)

(b) Define the following terms :

(i) Molality (m)

(ii) Abnormal molar mass

OR

(a) 30 g of urea (M = 60g mol^–1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg.

(b) Write two differences between ideal solutions and non-ideal solutions.

(a) Given:

Freezing point of sucrose (T_f) = 269.15K

Freezing point of water (T_f°) = 273.15K

Molar mass of sucrose = 342 gmol^-1

Molar mass of glucose = 180 gmol^-1

Freezing point of glucose (T_f) = ?

To calculate the freezing point of 10% glucose in water, first we need to calculate K_f , hence we apply the formula

Δ T_f = K_f × m

Where ΔT_f = Depression in freezing point

K_f = It is a constant known as molal depression constant

m = molality of the solution (the number of moles of the solute dissolved in 1000gms (1kg) of the solvent)

For sucrose:

First we will calculate ΔT_f, by applying the formula given below:

ΔT_f = T_f° - T_f

T_f° = freezing point of the pure solvent

T_f = freezing point of the solution

∴ ΔT_f = 273.15 – 269.15

⇒ ΔT_{f =} 4

Now, we will calculate m, by applying the formula given below:

As mass of solute (w₂) = 10 (given)

Molar mass of solute (M₂) = 342 (given)

Mass of the solvent in (M₁) = 100-10 = 90

Now, we will calculate K_f, by applying the formula

Δ T_f = K_f × m

K_f =

As m = 0.325 and ΔT_{f =} 4 (calculated above)

∴ K_f =

⇒ K_f = 12.30

For glucose:

To calculate the freezing point of 10% glucose in water, we apply the formula:

Δ T_f = K_f × m

As mass of solute (w₂) = 10 (given)

Molar mass of solute (M₂) = 180 (given)

Mass of the solvent in (M₁) = 100-10 = 90

K_f = 12.30 (calculated above)

Now, ΔT_f = ΔT_solvent – ΔT_solution

⇒ ΔT_f = 273.15 – 7.7

⇒ ΔT_f = 265.45 K

Thus, the freezing point of 10% glucose in water is 265.45 K

(b) (i) Molality = Molality of a solution is defined as the number of moles of the solute dissolved in 1000 grams (1kg) of the solvent. It is denoted by ‘m’.

Mathematically, m=

Or,

m= × 1000

Note: Molality is preferred when studies are made independent of temperature. This is because molality involves masses which do not change with temperature.

(ii) Abnormal Molar mass: For substances undergoing association, dissociation etc. in the solution, molecular masses determined from colligative properties is found to be different from expected value. These are called abnormal molecular masses.

Abnormal molecular masses are observed in any one of following cases:

⇒ When the solution is non-ideal, i.e., the solution is not dilute.

⇒ When the solute undergoes association in the solution.

⇒ When the solute undergoes dissociation in the solution.

Note: Colligative properties are those properties of ideal solution which depend only on the number of particles of the solute (molecules or ions) dissolved in a definite amount of the solvent and do not depend upon the nature of the solute.

OR

(a) Given:

Mass of Urea (W_urea) = 30 g

Molar mass of urea (M_urea) = 60 gmol^-1

Mass of water (W_H2O) = 846 g

Vapour pressure of the pure water(P°) = 23.8 mm Hg

Vapour pressure of water for the given solution = ?

To find out the vapour pressure of the solution, we apply the formula given below:

=

Where n₂ = no. of moles of the solute(urea) =

n₁ = no. of moles of the solvent (water) = =

∴ =

As P°= 23.8 (given)

W_urea= 30 g (given)

M_urea = 60 gmol^-1 (given)

W_H2O = 846 g (given)

M_H2O = 2× 1 + 16 = 18

∴ =

⇒ P_solution = 23.55 mm Hg

Thus, vapour pressure of water for the solution is 23.55 mm Hg.

(b) Differences between ideal solutions and non-ideal solutions are given below:

Note: Raoult’s Law states that “ In a solution, the vapour pressure of a component at a given temperature is equal to the mole fraction of that component in the solution multiplied by the vapour pressure of that component in the pure state.

According to Raoult’s law:

p_A = x_Ap°_A and p_B = x_Bp°_B

where p_A and p_B are the partial vapour pressures of the liquids A and B

p°_A and p°_B are the vapour pressures of the liquids A and B in the pure state.