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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