Chemistry Relative Lowering of Vapour Pressure and Elevation of Boiling Point
Click for Only Video

### Topics Covered :

● Colligative Properties
● Relative Lowering of Vapour Pressure
● Elevation of Boiling Point

### Colligative Properties and Determination of Molar Mass :

=> The vapour pressure of solution decreases when a non-volatile solute is added to a volatile solvent.

=> Many properties of solutions are connected with this decrease of vapour pressure. These are :

(i) Relative lowering of vapour pressure of the solvent

(ii) Depression of freezing point of the solvent

(iii) Elevation of boiling point of the solvent

(iv) Osmotic pressure of the solution.

=> Colligative Properties : These are the properties which depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution (colligative : from Latin : co means together, ligare means to bind).

### Relative Lowering of Vapour Pressure

The vapour pressure of a solvent in solution is less than that of the pure solvent. Raoult established that the lowering of vapour pressure depends only on the concentration of the solute particles and it is independent of their identity.

A relation between vapour pressure of the solution, mole fraction and vapour pressure of the solvent is given as

color{red}(p_1 = x_1p_1^0) .............(1)

The reduction in the vapour pressure of solvent (Δp_1) is given as :

color{red}(Δp_1 = p_1^0 – p_1 = p_1^0 - p_1^0 x_1)

color{red}(= p_1^0 (1-x_1)) ..............(2)

We know that color{red}(x_2 = 1 - x_1)

Equation (2) reduces to color{red}(Δp_1 = x_2 p_1^0) ...............(3).

For a solution containing several non-volatile solutes, the lowering of the vapour pressure depends on the sum of the mole fraction of different solutes. Equation can be written as

color{red}((Deltap_1)/(p_1^0) = ( p_1^0 - p_1)/p_1^0 = x_2) .....................(4).

The expression on the left hand side of the equation as mentioned earlier is called relative lowering of vapour pressure and is equal to the mole fraction of the solute. The above equation can be written as :

color{red}((p_1^0 - p_1)/(p_1^0) = n_2/(n_1+n_2) \ \ \ \ ( text(since) x_2 = n_2/(n_1+n_2))) ..................(5).

Here n_1 and n_2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions color{red}(n_2 < < n_1), hence neglecting n_2 in the denominator we have

color{red}((p_1^0 -p_1)/(p_1^0) = n_2/n_1) ...........................(6).

or color{red}((p_1^0 - p_1)/(p_1^0) = (w_2 xx M_1)/(M_2 xx w_1)) ..................(7).

Here w_1 and w_2 are the masses and M_1 and M_2 are the molar masses of the solvent and solute respectively.

From equation (7), the molar mass of solute (M_2) can be calculated.
Q 2906112078

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol^-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?
NCERT
Solution:

The various quantities known to us are as follows:

p_1^0 = 0.850 text(bar); p = 0.845 text(bar); M_1 = 78 g mol^–1; w_2 = 0.5 g; w_1 = 39 g

Substituting these values in equation (2.28), we get

(0.850 text(bar) – 0.845 text(bar))/(0.850 text(bar)) = (0.5 g xx 78 g mol^-1)/(M_2 xx 39 g)

Therefore, M_2 = 170 g mol^–1

### Elevation of Boiling Point :

=> The vapour pressure of a liquid increases with increase of temperature.

=> It boils at the temperature at which its vapour pressure is equal to the atmospheric pressure.

=> Example, water boils at 373.15 K (100°C) because at this temperature the vapour pressure of water is 1.013 bar (1 atmosphere).

=> Vapour pressure of the solvent decreases in the presence of non-volatile solute.

=> Variation of vapour pressure of the pure solvent and solution as a function of temperature is shown.

=> Example, the vapour pressure of an aqueous solution of sucrose is less than 1.013 bar at 373.15 K. In order to make this solution boil, its vapour pressure must be increased to 1.013 bar by raising the temperature above the boiling temperature of the pure solvent (water). Thus, the boiling point of a solution is always higher than that of the boiling point of the pure solvent in which the solution is prepared as shown in Fig. 2.7.

=> The elevation of boiling point also depends on the number of solute molecules rather than their nature.

Let T_b^0 be the boiling point of pure solvent and T_b be the boiling point of solution. The increase in the boiling point DeltaT_b = T_b - T_b^0 is known as elevation of boiling point.

Experimentally, for dilute solutions the elevation of boiling point (ΔT_b) is directly proportional to the molal concentration of the solute in a solution.

Thus, color{red}(DeltaT_b prop m)  .................(8)

or color{red}(DeltaT_b = K_b m) ............(9)

Here, the constant of proportionality K_b is called Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic Constant).

● The unit of K_b is K kg mol^(-1).

● Values of K_b for some common solvents are given in Table

=> If w_2 gram of solute of molar mass M_2 is dissolved in w_1 gram of solvent, then molality, m of the solution is given by the expression :

color{red}(m = ( w_2/M_2)/(w_1/1000) = ( 1000xx w_2)/(M_2 xx w_1)) .....................(10).

Substituting the value of molality in equation (2.30) we get color{red}(DeltaT_b = (K_b xx 1000 xx w_2)/(M_2 xx w_1)) .............(11).

color{red}(M_2 = (1000 xx w_2 xx K_b)/(DeltaT_b xx w_1)) ............(12).

Thus, for determining M_2, molar mass of the solute, known mass of solute in a known mass of the solvent is taken and ΔT_b is determined experimentally for a known solvent whose K_b value is known.

=> Value of K_b is given as :

color{red}(K_b = (R xx M_1 xx T_b^2)/(1000 xx Delta_text(vap) H)) ..............(13).

Here the symbols R and M_1 stand for the gas constant and molar mass of the solvent, respectively and T_b denote the boiling point of the pure solvent respectively in kelvin.

Further, Δ_text(vap)H represent the enthalpy of vapourisation of the solvent, respectively.
Q 2986134077

18 g of glucose, C_6H_(12)O_6, is dissolved in 1 kg of water in a saucepan. At what temperature will water boil at 1.013 bar? K_b for water is 0.52 K kg mol^(-1).

Solution:

Moles of glucose = 18 g// 180 g mol^(–1) = 0.1 mol
Number of kilograms of solvent = 1 kg
Thus molality of glucose solution = 0.1 mol kg^(-1)
For water, change in boiling point

DeltaT_b = Kb × m = 0.52 K kg mol^(–1) × 0.1 mol kg^(–1) = 0.052 K

Since water boils at 373.15 K at 1.013 bar pressure, therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K.
Q 2986234177

The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. K_b for benzene is 2.53 K kg mol^(–1)

Solution:

The elevation (DeltaT_b) in the boiling point = 354.11 K – 353. 23 K = 0.88 K

we know that M_2 = (2.53 K kg mol^(-1) xx 1.8 g xx 1000g kg^(-1))/(0.88K xx 90 g) = 58 g mol^(-1)

Therefore, molar mass of the solute M_2 = 58 g mol^(-1)