Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Given: Temperature = 373k

Vapour pressure of pure heptane (p⁰₁) = 105.2kpa and that of octane (p⁰₂) = 46.8kpa

Mass of heptane = 26 g

Mass of octane = 35 g

Molecular weight of heptane = C₇H₁₆ = 12 × 7 + 1 × 16 = 100 gmol^-1

Molecular weight of octane = C₈H₁₈ = 114 gmol^-1

Moles of heptane, n₁ = given mass /molecular weight = 26/100

⇒ n₁ = 0.26mol

Moles of octane, n₂ = given mass /molecular weight = 35/114

⇒ n₂ = 0.307mol

∴ Mole fraction of heptane,

⇒

⇒ χ₁ = 0.456

Now, Mole fraction of octane,

⇒

⇒ χ₂ = 0.544

∴ Partial pressure of heptane, p₁ = χ₁ × p⁰₁

⇒ p₁ = ₌ 0.456 × 105.2 = 47.97kpa

∴ Partial pressure of octane, p₁ = χ₂ × p⁰₂

⇒ p₂ = ₌ 0.544 × 46.8 = 25.46 kpa

∴ Total pressure exerted by solution = p₁ + p₂

= 47.97 + 25.46

= 73.43kpa