# 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

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