Heptane and octane form an ideal solution at 373 K. The vapour pressures of the pure liquids at this temperature are 105.2 kPa and 46.8 kPa, respectively. If the solution contains 25 g of heptane and 28.5 g of octane, calculate (i) vapour pressure exerted by heptane (ii) vapour pressure exerted by solution (iii) mole fraction of octane in the vapour phase.

Explanation

Heptane and octane form ideal solution
Mass of heptane = 25 g
Mol. mass of heptane ` = (7 xx 12) + (16 xx 1) = 84 + 16 = 100`
Mass of octane = 28.5 g
mol. mass of octane ` = (8 xx 12) + (1 xx 18) = 96 + 18 = 114`
Number of moles of heptane `(n_1) = 25/100 = 0.25`
Number of moles of octane `(n_2 ) = (28.5)/(114) = 0.25`
Mole fraction of heptane = `(n_1)/(n_1 + n_2) = (0.25)/(0.25 + 0.25) = (0.25)/(0.50) = 0.5`
Mole fraction of octane = 1 - 0.5 = 0.5.
(i) Vapour pressure exerted by heptane = 0.5 x 105.2 = 52.6 kPa
(ii) Vapour pressure exerted by octane = 0.5 x 46.8 = 23.4 kPa
Vapour pressure exerted by solution (heptane + octane) = 52.6 + 23.4 = 76 kPa
(iii) Mole fraction of uctane in the vapour phase = Vapour pressure of octane/Vapour pressure of solution
`= (23.4)/(76) = 0.3078`