# (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?

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Updated on: September 25th, 2023

(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

Total moles = 0.47 + 0.213 = 0.683 mol

XCH2Cl2 = 0.47/0.683 = 0.688 XCH2Cl2

XCHCl3 = 1.000 – 0.688 = 0.312

Ptotal = P1o + (P2o – P1o) 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.

Summary:-

(a) Vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 298 K are 200 mm Hg and 415 mm Hg respectively. 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.

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