● Colligative Properties

● Relative Lowering of Vapour Pressure

● Elevation of Boiling Point

● Relative Lowering of Vapour Pressure

● Elevation of Boiling Point

`=>` 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).

`=>` 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).

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.

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

NCERT

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`

`=>` 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.

`=>` 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)`.

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)`

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)`