A mixture of volatile organic solvents that is composed of 1.00 mole of heptane

Question

A mixture of volatile organic solvents that is composed of 1.00 mole of heptane (C_7 H_16) and 1.75 moles of octane (C_8 H_18) is found to have a vapor pressure of 53.3 torr at 25 degree C. The vapor pressure of pure heptane and octane at 25 degree C are 92.0 torr and 31.2 torr. respectively. (a) Calculate the mole fraction of heptane and octane in the mixture, (b) Calculate the partial vapor pressure of heptane and octane in the mixture, and determine whether heptane and octane form an ideal solution, (c) Calculate the mole fraction of octane in the vapor phase above this solution?

Explanation / Answer

Mole fraction of heptane =1/2.75 =0.3636

Mole fraction of octane=1.75/2.75=0.6363

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Here, the vapor pressure exerted by the mixture will be total pressure.

y(heptane)=(x*P sat)/P=(0.3636*92)/53.3=0.6276

y(octane )=(0.6363*31.2)/(53.3)=0.3724

Therefore, the mole fraction of heptane and octane in teh vapor phase are 0.6276 and 0.3724.

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Mole fraction of heptane in vapor phase (y hep)=partial pressure of heptane/total pressure

Partial pressure of heptane=(0.6276*53.3)=33.45 torr

Partial pressure of octane=(0.3724*53.3)=19.84 torr

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