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#vapor phase

5 public questions tagged with this topic.

A solution of two volatile liquids has vapor pressures of 250 mm Hg and 350 mm Hg for pure components. If the total vapo

Liquid phase: 310 = 250 x₁ + 350 (1 - x₁) . 310 = 250 x₁ + 350 - 350 x₁ , 100 x₁ = 40 , x₁ = 0.4 , x₂ = 0.6 . Vapor phase: y₁ = (P₁⁰ · x₁/Ptotal) = (250 × 0.4/310) ≈ 0.3226 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

A solution of two volatile liquids has vapor pressures of 600 mm Hg and 800 mm Hg. If the total vapor pressure is 680 mm

Liquid phase: 680 = 600 x₁ + 800 (1 - x₁) . 680 = 600 x₁ + 800 - 800 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₁ = (600 × 0.6/680) ≈ 0.5294 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

A solution of two volatile liquids has vapor pressures of 500 mm Hg and 700 mm Hg. If the total vapor pressure is 580 mm

Liquid phase: 580 = 500 x₁ + 700 (1 - x₁) . 580 = 500 x₁ + 700 - 700 x₁ , 200 x₁ = 120 , x₁ = 0.6 , x₂ = 0.4 . Vapor phase: y₁ = (500 × 0.6/580) ≈ 0.5172 .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Vapour Pressure and Raoult's Law - Ideal and Non-ideal Solutions

A solution of two volatile liquids has vapor pressures of 300 mm Hg and 400 mm Hg for pure components. If the mole fract

Vapor phase: y₂ = (P₂⁰ · x₂/Ptotal) . 0.6 = (400 · x₂/Ptotal) . Liquid phase: Ptotal = 300 (1 - x₂) + 400 x₂ . Substitute: 0.6 Ptotal = 400 x₂ , Ptotal = 300 + 100 x₂ . 0.6 (300 + 100 x₂) = 400 x₂ , 180 + 60 x₂ = 400 x₂ , 340 x₂ = 180 , x₂ = (180/340) ≈ 0.5294 . Ptotal = 300 (1 - 0.5294) + 400 × 0.5294 ≈ 352.96 mm Hg .

Ref: NCERT Class 12 Chemistry > Chapter 1: Solutions > Topic: Types of Solutions and Expressing Concentration