The Foundations of Chemistry

cule, and similarly for each octane molecule. The properties of such a solution can be
predicted from a knowledge of its composition and the properties of each component.
Such a solution is very nearly ideal.
Consider an ideal solution of two volatile components, A and B. The vapor pressure
of each component above the solution is proportional to its mole fraction in the solution.

PAXAP^0 A and PBXBP^0 B

The total vapor pressure of the solution is, by Dalton’s Law of Partial Pressures (Section
12-11), equal to the sum of the vapor pressures of the two components.

PtotalPA
PB or PtotalXAP^0 A
XBP^0 B

This is shown graphically in Figure 14-10. We can use these relationships to predict the
vapor pressures of an ideal solution, as Example 14-5 illustrates.

EXAMPLE 14-5 Vapor Pressure of a Solution of Volatile Components

At 40°C, the vapor pressure of pure heptane is 92.0 torr and the vapor pressure of pure octane
is 31.0 torr. Consider a solution that contains 1.00 mole of heptane and 4.00 moles of octane.
Calculate the vapor pressure of each component and the total vapor pressure above the solution.

Plan

We first calculate the mole fraction of each component in the liquid solution. Then we apply
Raoult’s Law to each of the two volatile components. The total vapor pressure is the sum of
the vapor pressures of the components.

Solution

We first calculate the mole fraction of each component in the liquid solution.

Xheptane0.200

Xoctane 1 Xheptane0.800

Then, applying Raoult’s Law for volatile components,

PheptaneXheptaneP^0 heptane(0.200)(92.0 torr) 18.4 torr

PoctaneXoctaneP^0 octane(0.800)(31.0 torr) 24.8 torr

PtotalPheptane
Poctane18.4 torr
24.8 torr 43.2 torr

You should now work Exercise 40.

The vapor in equilibrium with a liquid solution of two or more volatile components
has a higher mole fraction of the more volatile component than does the liquid solution.

EXAMPLE 14-6 Composition of Vapor

Calculate the mole fractions of heptane and octane in the vapor that is in equilibrium with the
solution in Example 14-5.

Plan

We learned in Section 12-11 that the mole fraction of a component in a gaseous mixture equals
the ratio of its partial pressure to the total pressure. In Example 14-5 we calculated the partial
pressure of each component in the vapor and the total vapor pressure.

1.00 mol heptane



(1.00 mol heptane)
(4.00 mol octane)

If component B were nonvolatile,

then P^0 Bwould be zero, and this

description would be the same as

that given earlier for a solution of a

nonvolatile, nonionizing solute in a

volatile solvent, PtotalPsolvent.

14- 9Lowering of Vapor Pressure and Raoult’s Law 55 9

Figure 14-10 Raoult’s Law for

an ideal solution of two volatile

components. The left-hand side

of the plot corresponds to pure B

(XA0, XB1), and the right-hand

side corresponds to pure A (XA1,

XB0). Of these hypothetical

liquids, B is more volatile than A

(P^0 B P^0 A).

Vapor pressure

P^0 B

0

10

0

0 1

XB

XA

P^0 A

Ptotal = PA + PB

PB

PA