Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°c. assuming concentration of solute is much lower than the concentration of solvent, determine the vapour pressure (mm of hg) of the solution.

Answer 1

Answer:

Weight of non-volatile non-electrolyte solute = 2.5 g

Weight of the solvent = 100 g

The elevation in boiling point, ∆T = 2 ℃  

We know, the elevation of boiling point is given by  

∆T = m * Kb

Where m = molality = (weight of solute)/(molecular weight of solute)*(1000/weight of solvent)

Kb = boiling point elevation constant = 0.76 K kg mol⁻¹

∴ ∆T = m * Kb

Or, 2 = [2.5 / molecular weight of solute]* 1000/100 * 0.76

Or, molecular weight of solute = 9.5 g

Now, by using Raoult’s Law, we have

[P° - P] /P° = nsolute / nsolvent ……[∵ nsolvent >>> nsolute ∴ nsolution = nsolvent]

Where P = vapour pressure of the solution  and P° = vapour pressure of the solvent = 1atm = 760 mmHg

∴ [P° - P] /P° = nsolute / nsolvent

Or, [P° - P] /P° = [(weight of solute)/(molecular weight of solute)] / [(weight of solvent)/(molecular weight of solvent)]

Or, [760 – P] / 760 = [2.5/9.5] / [100/18] …… [ Molecular weight of water(solvent) = 18 g]

Or, 760 – P = 35.981

Or, P = 760 – 35.81 = 724 mmHg

Hence, the vapour pressure of the solution is 724 mmHg.

Hope this helps!!!!!