1. Aim # 46: How does a solute affect the vapor pressure of a solvent?
H.W. # 46
Study pp. 521 – 526(sec. 11.4)
Ans. ques. p. 521 # 51,54,59,64

2. pure solvent solution with a nonvolatile solute Vsolvent = Vsolution
I Vapor pressure is the pressure exerted by vapor that is in dynamic equilibrium with its liquid (in a closed system).
pure solvent solution with a nonvolatile solute Vsolvent = Vsolution

3. II Raoult’s law. Where. PA = vapor pressure of the
II Raoult’s law Where PA = vapor pressure of the PA = χ PA solution (solvent) χA = mole fraction of the solvent PA0 = vapor pressure of the pure solvent
vapor pressure of the pure solvent
solution
vapor
pressure PA
mole fraction of solvent χA Vapor pressure depression is proportional to the concentration of solute particles.

4. Problem: Urea, (NH3)2CO, has practically no vapor pressure at ordinary temperatures. The vapor pressure of pure water at 30.0C is torr. A solution contains 50.0 g urea and g H2O. Find the vapor pressure of this solution at 30.0C.
Ans: nurea = 50.0 g x 1mol = .833 mol g nH20 = g x 1 mol = mol g PH20 = χH20P0H PH20 = x torr = torr
The vapor pressure is lower by torr.

5. Problem: The vapor pressure of pure ethanol at 63. 50C is 400. torr
Problem: The vapor pressure of pure ethanol at 63.50C is 400. torr. When 15 g of an unknown, non-volatile solute is dissolved in g of ethanol, the vapor pressure of the resulting solution is 380 torr. What is the molar mass of the solute?
Ans: PEtOH = χEtOHP0EtOH χEtOH = PEtOH = 380 torr = P0EtOH torr nEtOH = x 1mol = mol g
ntotal = = mol
nsolute = – = .105 mol molar mass = g ͘ mol mol
molar mass = g = 140 g

6. III The vapor pressure of a solution with an ionic solute
vapor pressure α concentration depression of solute
In solution NaCl(aq) → Na+(aq) + Cl-(aq)
For a 1.0 M NaCl solution, the number of moles/liter of solute particles is actually For glucose(1.0M) For NaCl(1.0M) XH2O = nH2O χH2O = nH2O nglucose + nH2O nNaCl + nH2O
Since PA = χAP0A , the ionic solute depresses the vapor pressure more.

7. Forces between Deviation
IV Solutions that obey Raoult’s law exactly are called ideal solutions.
Forces between Deviation
solvent(A) and solute(B) from Raoult’s particles law example
A↔A, B↔B ≈ A↔B none(ideal hexane solution) pentane
A↔A, B↔B < A↔B PA↓ ethanol (neg. deviation) water
A↔A, B↔B > A↔B PA↑ ethanol (pos. deviation) hexane

9. V Solutions of two volatile liquids (e.g. n-butane and n-pentane)
Solutions whose vapor pressures deviate from Raoult’s law are nonideal solutions.
V Solutions of two volatile liquids (e.g. n-butane and n-pentane)
Total vapor pressure = sum of the partial pressures PT = PA + PB = χAPA0 + χPB0
Problem: At 250c, the vapor pressure of n-butane,
C4H10, is 1823 torr. The vapor pressure of
n-pentane, C5H12, is 521 torr. Assume that
solutions of these two compounds are ideal.

10. Ans: 10.0 g n-butane x 1 mol = .172 mol 58.12 g
a) Find the total vapor pressure at 250c of a g solution containing % n-butane and 90% n-pentane (m/m).
Ans: g n-butane x 1 mol = mol g
90.0 g n-pentane x 1 mol = mol g
χn-butane = =
χn-pentane = =
PT = (.121)(1823 torr) + (.879)(521 torr) PT = 679 torr

11. b) What is the mole fraction of the butane in the vapor?
Ans: From Dalton’s law of partial pressures, Pn-butane = nn-butane = (.121)(1823 torr) PT nT torr = .325
c) Find the mole fraction of n-butane in a n-butane-n-pentane solution having a total vapor pressure of 760 torr at 250C Hint: χn-pentane = 1 – χn-butane

12. 760 = χbPb0 + (1 – χb)Pp0 Ans: PT = 760 = χbPb0 + χpPp0
760 = χbPb0 + Pp0 – χbPp0
760 = χb(1823) – χb(521)
solving for χb ,
χb = 760 – 521 = – 521
Additional Problems:
Zumdahl (8th ed.) p. 523 # 99
p. 521 # 52, 57, 58, 60, 63