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 ?

Explanation

Molar mass of heptane `(C_7H_16)= 100 g "mol"^(-1)`
Molar mass of octane `(C_8H_18) = 114 g "mol"^(-1)`
Number of moles in 26.0 g heptane = `(26.0)/(100 g "mol"^(-1)) = 0.26 `mol
Number of moles in 35.0 g octane = `(35.0)/(114 g "mol"^(-1) ) = 0.31 `mol
Mole fraction of heptane ` = (0.26)/(0.26 + 0.31) = 0.456`
Mole fraction of octane = 1 - 0.456 = 0.544
Partial pressure of heptane = 0.456 x 105.2 kPa = 47.97 kPa
Partial pressure of octane = 0.544 x 46.8 kPa = 25.46 kPa
Total pressure = 47.97 + 25.46 = 73.43 kPa.