Question

The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in a molar ratio of 3 : 2 is

[At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

Options

• 350 mm of Hg

• 160 mm of Hg

• 168 mm of Hg

• 336 mm of Hg

Answer 1

336 mm of Hg

Explanation:

Given: Molar ratio → Benzene : Octane = 3 : 2

⇒ Total moles = 5

⇒ Mole fraction of benzene `chi_"benzene" = 3/5 = 0.6`

⇒ Mole fraction of octane `chi_"octane" = 2/5 = 0.4`

Vapour pressure at 45°C

`P_"benzene"^circ = 280` mm Hg

`P_"octane"^circ = 420` mm Hg

By using Raoult’s law for ideal solutions

`P_"solution" = chi_"benzene" * P_"benzene"^circ + chi_"octane" * P_"octane"^circ`

= (0.6 × 280) + (0.4 × 420)

= 168 + 168

= 336 mm Hg