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E 100. Two liquids, A and B, have vapor a given tempera ture of 24 and mmHg, res

ID: 520156 • Letter: E

Question

E 100. Two liquids, A and B, have vapor a given tempera ture of 24 and mmHg, respectively. prepare tions of A and B at a given temperature and measure the total pressures above the solutions. We obtain this data: Solution Amt A (mol) Amt B (mol) P (mmHg) 32 Predict the total pressure above a solution of 5 mol A and 1 mol B. 101. Three 1.0-L flasks other with stopco the flasks contain third, 0.50 g of 308 K is 42 mml cocks are then o pressure? 102. Butane (C4H10) normal boiling 0.55 g of butan liquid? If the bu a liquid?

Explanation / Answer

100)

From the given data the mixture of liquids A and B is an ideal solution and follows Raoult's law

Thus

P total = Xa .Pa0 + Xb.Pb0

Given .Pa0 = 24 and .Pb0 = 36 mm

Thus when 5 moles of A and 1 mole of B are mixed

Xa = 5/6 and mole fraction of B =Xb = 1/6

Ptotal = (5/6) 24 + (1/6) 36

= 26mm/Hg

101)

Three gases in 3 1 L flasks.

When they are mixed by opening the stop cocks , the total volume of container = 3L

The new partial pressure of gases at 3L can be calted by using P1V1 = P2V2

New Partial pressure of N2 = 1L x 1 atm /3L = 1/3atm

= 760/3 mm

new pressure of water = 42mm {saturated vapor pressure of water does not change with volume, but with temperautre]

new pressure of ethanol = 1L x 102 /3L

= 102/3 mm/Hg

Thus total pressure = (760/3 ) + 42 + (102/3)

= 329.33 mmHg