What will be the partial vapor pressure of hexane at 68oC above the mixture of cyclohexane - hexane that contains 34 % of hexane by volume if it is known that the vapor pressure of pure hexane at the given temperature is 760 mm Hg

the vapor pressure will be pV = 298.4 mm Hg

Explanation:

Assuming ideal behaviour of liquid mixture of the cyclohexane - hexane, then the Raoult equation applies:

pL = p⁰*x

where pL = partial pressure of the liquid in the mixture, p⁰ = partial pressure of pure hexane, and x = partial molar fraction of hexane in the mixture = 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 = p⁰*x = 760 mm Hg * 0.34 = 298.4 mm Hg

pV = 298.4 mm Hg