Answer:
the vapor pressure will be pV= Ā 298.4 mm Hg
Explanation:
Assuming ideal behaviour of liquid mixture of the cyclohexane - hexane, then the Raoult equation applies:
pL= pā°*x
where pL = partial pressure of the liquid in the mixture Ā , pā°= partial pressure of pure hexane Ā , and Ā x= partial molar fraction of hexane in the mixture = Ā 0.34
(if we assume that they have approximately the same density , 34 (V/V)% = 0.34 of partial molar fraction of hexane )
Then since the liquid mixture is at equilibrium with the vapor , the partial vapor pressure pV should be equal to the liquid vapor pressure. Thus
pV=pL
pV= pā°*x = 760 mm Hg* 0.34 = 298.4 mm Hg
pV= Ā 298.4 mm Hg
