(a) Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are 200 mm Hg and 415 mm Hg respectively. Calculate the vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at 298K (b) State Raoult’s law (c) What is meant by azeotropic mixture?

(a) The vapour pressure of the solution prepared by mixing 25.5 g of CHCl3 and 40 g of CH₂Cl₂ at 298K is 347.9 mm of Hg

(b) For any solution the partial vapour pressure of each volatile compound in the solution is directly proportional to the mole fraction is Raoult’s law.

(c) Binary mixtures having the same composition in liquid and vapour phase and boil at constant temperature are azeotropic mixture.

(a) Given that VP of CHCl₃ = 200 mm of Hg

And VP of CH₂Cl₂ = 415 mm of Hg

We know that Molar mass of CH₂Cl₂ = 85

Molar mass of CHCl₃ = 119.5

Now consider Moles of CH₂Cl₂ = 40/85 = 0.47

Moles of CHCl₃ = 25.5/119.5 = 0.213

Therefore adding them we get

Total moles = 0.47 + 0.213 = 0.683 mol

X_CH2Cl2 = 0.47/0.683 = 0.688 X_CH2Cl2

X_CHCl3 = 1.000 – 0.688 = 0.312

P_total = P₁^o + (P₂^o – P₁^o) X₂

= 200 + (415 – 200) x 0.688

On simplifying we get

= 200 + 147.9

= 347.9 mm of Hg

(b) Raoult’s law – Each volatile substance in a solution has a partial vapour pressure that is precisely proportional to the mole fraction.

(c) Azeotropes: binary mixures that boil at a constant temperature and have the same composition in the liquid and vapour phases.