Question

vapour pressure of an non volatile solute is 730 mm Hg at 100 degree centigrate . The boiling point of aqueous solution will be

(a) 100.139 *c
b 102.19*c
c. 100.114*c
d 100.22*c
solve with solution

Answer 1

Decrease in the vapour pressure is directly proportional to the concentration x (mole fraction) of the solute in water.

Δp = x * P
(760-730) = x * 760          =>     x = 30/760 
moles of solute in 1 kg of water = 30/760 * (1000/18) = 2.193

Boiling point elevation ΔT = K_b * i * b_solute
      K_b = Ebullioscopic constant for water = 0.512 °K/ Molal
             i = 1 assumed as the value is not given.
       b_solute = moles/kg of solute

ΔT = 0.512° K  / (moles/kg) * b_solute 
      = 0.512°K * 2.193 = 1.122°K

So the boiling point is = 100+1.122° C = 101.122°C

If the solute is ionic in solution like salt, then i = 2 in the Raoult's formula above.

Hence,  ΔT = 2.244°C

Then Boiling point = 102.244°C