A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be :

Question

A solution has a 1: 4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20°C are 440 mm of Hg for pentane and 120 mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be : (A) 0.178 (B) 0.278 (C) 0.378 (D) 0.478

Tardigrade Admin · Accepted Answer

Total vapour pressure of mixture = (Mole fraction of pentane V.P. of pentane) + (Mole fraction of hexane x V.P. of hexane) = V.P. of pentane in mixture + V.P. of hexane in mixture

= (\frac{1}{5} \times 440 + \frac{4}{5} \times 120) = 184 mm

∵

V.P. of pentane in mixture = V.P. of mixture x mole fraction of pentane it vapour phase

88 = 184

x mole fraction of pentane in vapour phase

∴

Mole fraction of pentane in vapour phase

= \frac{88}{184} = 0.478

Q. A solution has a 1:4 mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at 20∘C are 440mm of Hg for pentane and 120mm of Hg for hexane. The mole fraction of pentane in the vapour phase would be :

0.178

0.278

0.378

0.478

Q. A solution has a $1 : 4$ mole ratio of pentane to hexane. The vapour pressure of the pure hydrocarbons at $2 0^{\circ} C$ are $440 mm$ of $H g$ for pentane and $120 mm$ of $H g$ for hexane. The mole fraction of pentane in the vapour phase would be :