Refer to MtC figure 3-17 to answer: Note that each vertical line in the graph gr

Question

Refer to MtC figure 3-17 to answer:

Note that each vertical line in the graph grid equals to 5%, and the scale runs both ways for benzene and toluene:

100% Benzene ----- 0% Benzene

0% Toluene--------- 100% Toluene

(a) What is the % composition of the vapor in equilibrium with a boiling liquid that has a composition of 30% benzene and 70% toluene.

(b) If a sample of vapor is found to have a composition of equal parts benzene and toluene, what is the composition of the boiling liquid that produced this vapor.

Explanation / Answer

a.) Calculate the mole fraction of each substance:

Benzene 30% = 0.3
Toluene = 70% = 0.7

Use Raoult's Law:

Psolution = P°benzene benzene + P°toluene toluene

Psolution = (94.6 torr) (0.333) + (29.1 torr) (0.667) = 44.289 torr + 7.604 torr = 50.9115torr

I won't round off here because I want to go to the mole fractions in the vapor.

3) Calculate mole fractions in vapor:

Benzene(94.6 torr) (0.333) / 51.893 = 31.5018 / 51.893 = 0.607
Toluene (29.1 torr) (0.667) / 51.893 =19.4097 / 51.893 = 0.374

Hence, there is 60.7% of benzene and 37.4% of toluene in vapour.

b.) Similarly the process can be done in reverse manner the vapour compostion is given and we have to find the compostion. Just go the steps in reverse order.

The precent composition will have 51% benzene and 49% toluene.

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