Chem104 EXAM 2Av1 SPRING, 2018 RAVE At 80°C the vapor pressures of pure liquid b

Question

Chem104 EXAM 2Av1 SPRING, 2018 RAVE At 80°C the vapor pressures of pure liquid benzene and toluene are 745 torr and 290 torr, respectively. A solution prepared by mixing benzene and toluene obeys Raoult's Law. At this temperature, if the mole fraction of benzene in the solution is 0.340, determine; a. (3 pts) the partial pressure of toluene in the vapor above the solution. b. (3 pts) the total pressure of the vapor above the solution. C. (3 pts) the mole fraction of toluene in the vapor above the solution is? 25. (3 pts) A 5.50 gram sample of a non volatile nonelectrolyte is dissolved in 250 grams of benzene. The freezing point of this solution is 1.02°C below that of pure benzene. What is the molar mass of the unknown compound? (K, -5.12o9) oC. mole

Explanation / Answer

1. Given P°(toluene) = 290 torr; mole fraction, x(toluene) = 0.34

Partial pressure of toluene, P(toluene) = x(toluene) × P°

= (1-0.34) × 290 torr = 191.4 torr

2. Total vapour pressure = P°(tol)*x(tol) + P°(benzene)*x(benzene)

= (1-0.34)*290 torr + 0.34*745torr = 444.7

3. Mole fraction of toluene in vapour phase is given by

= (Vapour pressure of toluene)/(total vapour pressure)

= 191.4/444.7 = 0.43

4. Depression in freezing point is given by, ?Tf = m × Kf,

m = molality and Kf = freezing point depression constant.

Given ?Tf = 1.02°C, Kf = 5.12(°C kg) / mole

Substituting, m = ?Tf/Kf = 1.02°C/5.12(°C kg/mole)

m = 0.1992 mole/kg

Hence 1000kg of solvent has = 0.1992 moles

250g of solvent has = (0.1992×250)/1000 = 0.0498moles

Now, 5.5 g of solute was dissolved.

Thus 5.5 g of solute has 0.0498 moles

Thus molecular weight of unknown = 5.5/0.0498 = 110.44g

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