Elevation in boiling point

Elevation in boiling point-

source : sciencing

Elevation in boiling point-

        Boiling point of a liquid is a characteristic temperature at which its vapour pressure becomes equal to one atmosphere or 760 mm or 76 cm pressure(atmospheric pressure).

       Addition of a nonvolatile solute in a solvent lowers its vapour pressure and thus more heat is required to increase the vapour pressure of solution upto one atmosphere.Hence the boiling point of the solvent is elevated when nonvolatile solute is added. It means boiling point of solution is greater than boiling point of  pure solvent.This increase in boiling point is known as elevation in boiling point or the difference in boiling point of the solution and pure solvent is termed elevation in boiling point .

source : socratic.org

source : edu.glogster.com

Elevation in boiling point = ΔTb = boiling point of solution -boiling point of pure solvent

            When a graph is plotted between vapour pressure and temperature for a pure solvent and two solutions of different concentrations. The curves of the solutions always lie below the curve of the pure solvent.

source : eXe

P_o=vapour pressure of pure solvent

P₁= vapour pressure of solution I

P₂= vapour pressure of solution II

T₀=boiling point of pure solvent

T₁= boiling point of solution I

T₂= boiling point of solution II

Line P₀ I represents the atmospheric pressure

If solutions are very dilute then these curves may be taken as straight  lines near the boiling point .

Hence ,

ΔGIK and  ΔGHJ are similar

So,

GI/GH =GK/GJ

(T₂-T₀) / (T₁-T₀) = (P₀-P₂) / (P₀-P₁)

ΔT₂ /ΔT₁= ΔP₂ /ΔP₁

ΔT_b ∝  ΔP

According to Raoult’s law,

(p₀-p_s) / p₀ = wM/Wm

(p₀-p_s)  = [wM/Wm] x p₀

p₀ (vapour pressure of pure solvent) and molecular weight of solvent (M) are constant.

So,

(p₀-p_s)  = w / mW

Δp  ∝  w/ m W

Δp ∝  ΔTb ∝  w/mW

Thus ,

ΔT_b   ∝  w/ m W

ΔT_b  = K w/ m W

K = Elevation constant

when w/m = 1 mole, W= 1 gram then,

ΔT_b = K

when w/m = 1 mole, W= 100 gram then,

ΔT_b = K /100 = K_b‘

K_b‘ is molecular or molar elevation constant

ΔT_b  =  100 K_b‘ w/m W

Definition of Molecular or molar elevation constant-

If w/m = 1 mole, W =100 gram then ΔT_b = K_b‘

            ” Molar elevation constant is equal to the elevation in boiling point which is produced when  one mole of nonvolatile ,non electrolyte solute is dissolved in 100 gram of solvent.

Definition of  molal elevation constant-

If w/m = 1 mole, W =1000 gram then ΔT_b = K_b

          ” Molal elevation constant is equal to the elevation in boiling point which is produced when  one mole of nonvolatile ,non electrolyte solute is dissolved in 1000 gram of solvent.

We know,

w x 1000/m W = molality

So,

ΔT_b = molality x K_b

Molal elevation constant can be calculated as ,

K_b = R T_b²/1000 L_v

R = gas constant (in calorie), Tb= boiling point of solvent in Kelvin

L_v = Latent heat of vaporisation (calorie /gram)

IMPORTANT POINTS TO REMEMBER-

1) Elevation of boiling point is a colligative property because it depends upon the molality of solution (means number of moles present in definite amount of solvent).

2) Equimolal solutions of different nonvolatile and non electrolyte substances dissolved in the same solvent show the same elevation in boiling point (colligative property).