4 15. If 0.35 moles of A and 0.45 moles of B are steam distilled, what is the to

Question

4 15. If 0.35 moles of A and 0.45 moles of B are steam distilled, what is the total vapot pressure of the solution at 90°c? (SHOW WORK) (4) 16. It the external pressure is 700 torr, at what temperature specifically (give or take 1°C) will A& water codistill? EXPLAIN. (6) 17. At 99.6, w ater has a vapor pressure of 750 torr and quinoline (MW 129) has a vapor pressure of 10 torr. What weight of water must be distilled for each gram of quinoline in a steam distillation at 760 torr? SHOW WORK ON BACK SIDE OF THIS PAGE.

Explanation / Answer

According to Raoult's Law

we need to calculate the mole fraction = (moles of B)/(moles of A + Moles of B)

=0.45/(0.35+0.45)

= 0.5625

As from the provided graph, we could see the vapor pressure at 90 degree C =160

so total pressure = 0.5625 × 160 = 90 torr

c) P = Pm  × Xa

Xa is mole fraction in gaseous phase

moleculae weight of quinolene =129

mole of quinolene = w2/m2 = w2 /129

moles of water = w1/m1 = w1/18

total moles = (w1/18 )+ (w2/129)

Xquinolene = moles of quinolene/ total moles

Xwater = mole of water /total moles

For quinolene

P =Pm × Xquinolene

10 = 760 × (w2/129) / [(w1/18)+(w2/129) ] -------1

for water

P = Pm × Xwater

750 = 760  × (w2/18) / [(w1/18)+(w2/129) ] -------2

dividing equation 1 and 2, we get

10/750 = (w2/129)/ (w1/18)

given w2 = 1 gm

1/75 = 18/(w1× 129)

w1 = 10.46 gm

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