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) 0.549 (B) 0.200 (C) 0.786 (D) 0.478

Tardigrade Admin · Accepted Answer

Total vapour pressure of mixture = (Mole fraction of pentane × VP of pentane) + (Mole fraction of hexane × VP of hexane) = VP of pentane in mixture + VP of hexane in mixture. =((1/5) × 440+(4/5) × 120)=184 mm ∵ VP of Pentane in mixture. = VP of mixture × mole fraction of pentane in vapour phase 88=184 × mole fraction of pentane in vapour phase ∴ Mole fraction of pentane in vapour phase =(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

0.549

0.200

0.786

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