Tried this problem for at least 45 mins; still dont get it :( Benzene (C6H6) and

Question

Tried this problem for at least 45 mins; still dont get it :(

Benzene (C6H6) and toluene (C7H8) form ideal solution that obeys Raoult's law. At 40 degree C. the vapor pressure of benzene is 180 torr and that of toluene is 60 torr. (a) Which liquid has the higher boiling point? Explain (b) What will be, the total vapor pressure (in torr) of a solution prepared by mixing equal masses of the two liquids? (c) Suppose you wished to prepare a solution that will have a total vapor pressure of 96 torr at 40 degree C. What must be the mole percent concentrations (or mole fraction) of each liquid in the solution?

Explanation / Answer

for the second bit of the question that is
b) p = paxa + pbxb where Pa and Pb are the vapour pressures and Xa and Xb are the mole fractions respectively which is 1/2 in this case.

So Pa = vapour pressure of benzene = 180 torr

and Pb = vapour pressure of toluene = 60 torr

p = 180 * 0.5 + 60 * 0.5

= 120 torr

c) p =paxa + pbxb which can be written as paxa + pb(1-Xa)

here

Pa = vapour pressure of benzene = 180 torr

and Pb = vapour pressure of toluene = 60 torr and p = 96 torr

so we need to find out Xa

96 = 180Xa + 90(1-Xa)

=> solving we get Xa = mole fraction of benzene = 1/15

so Xb = mole fraction of toluene = 14/15

a) the boiling point of toluene is more because it has less vapour pressure and the Boiling point shares an inverse relation with the vapour pressure.

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