Question

A) Calculate the vapour pressure of a 5% by mass benzoic acid [C₇H₆O_2(aq)] in ethanol solution at 35^oC. The vapour pressure of pure ethanol at this temperature is 13.40 kPa. Present your answer in kPa.

Answer 1

A 5% BY MASS Benzoic acid in Ethanol solution means 5gm of Benzoic acid is present in 100 gm of Ethanol.

So ) weight of solute (Benzoic acid ) W₁ = 5 gm

weight of solvent(Ethanol) W 2 = 100gm

) no.of moles of solute = weight of benzoic acid /molecular weight of benzoic acid =

no.of moles of solvent =  weight ofEthanol /molecular weight of Ethanol =

) mole fraction of Benzoic acid =

  

  

This problem belongs to relative lowering of vapour pressure and we find out vapour pressure of the solution by following equation

  

that means relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute .

so in the problem given info. is

P_{0 =} vapour pressure of the pure solven (Ethanol) = 13.40 k Pa

P = vapour pressure of solution

P₀ --P/ P₀ = nB /nB+ nE

13.4-P/13.4 = 0.04/2.21

P = 13.40-0.2412

=13.1588 k Pa

So the vapour pressure of the solution is 13.1588