The vapour pressure of pure benzene at 20 °C is 75 Torr and that of pure methylbenzene is 25 Torr at the same tempera-ture. In a

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The vapour pressure of pure benzene at 20 °C is 75 Torr and that of pure methylbenzene is 25 Torr at the same tempera-ture. In a

n equimolar mixture xbenzene = xmethylbenzene = 1 so the 2 partial vapour pressure of each one in the mixture is_________.

Chemistry

1 answer:

ss7ja [257] · Answer 1 · 2022-05-23T16:00:16+03:00

Answer:

Benzene: 37.5 Torr

Methylbenzene: 12.5 Torr

Explanation:

By Raoult's Law, each substance in a gas mixture contributes to the total pressure of the mixture proportionally to their respective mole fraction. So,

Ppartial = x*P°

Where x is the mole fraction (0.5 for each one because it's equimolar), and P° is the vapor pressure.

Benzene: Ppartial = 0.5 * 75 = 37.5 Torr

Methylbenzene: Ppartial = 0.5 * 25 = 12.5 Torr