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

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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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