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Raoult's Law

>> Parts of this equation/concept include:

A -	Total Vapor Pressure
B -	Purity from Distillation

Raoult's law states that the vapor pressure due to a volatile component of the system (P_A) is

P_A = X_AP_A°

where X_A is the mole fraction of A and X_A° is the vapor pressure of pure A. The pure vapor pressure of water can be determined from Table 8.2. Other vapor pressures will be given as part of the problem.

Dalton's law (Chapter 8) says that the total pressure is the sum of the vapor pressures. So to determine the total vapor pressure, determine the partial pressure for each component (from Raoult's Law) and add them together.

>> Example 1

In a mixture of 86.0 g C₆H₆ (P° = 93.96 torr) and 90.0 g C₂H₄Cl₂ (P° = 224.9 torr), what is the total vapor pressure?

Solution:

Since this is a mixture, Raoult's law applies. Since both substances are volatile (both have a vapor pressure), the partial pressure of each must be determined.

Raoult's law = P_A = X_AP_A°

First, determine the mole fraction for each substance.

moles of benzene (B), C6H6:	86.0 g B		1 mol 78.11 g	=	1.10 mol
moles of dichloroethane (D), C2H4Cl2:	90.0 g D		1 mol 98.94 g	=	0.910 mol

XB =

1.10 1.10 + 0.910

=

0.547

XD =

0.910 1.10 + 0.910

=

1 – 0.547 = 0.453

Mole fraction has no units.

partial pressure of benzene (PB)	=	XBPB°
	=	(0.547)(93.96 torr)
	=	51.2 torr

partial pressure of dichloroethane (PD)	=	XDPD°
	=	(0.453)(224.9)
	=	102 torr

The total pressure is the sum of the partial pressures.

Since you are adding, count decimal places to determine significant figures.

P_T = P_B + P_D = 51.2 + 102 = 153 torr

Note that the answer has a value between the two values of pure vapor pressures. This will always be the case. The value will be closest to the vapor pressure of the substance with the highest mole fraction. This is a good way to check your work.

>> Example 2

What is the total vapor pressure in a mixture of 50.0 g CH₃OH (P° = 93.3 torr) and 25.0 g H₂O (P° = 17.5 torr)?

Solution:

First, determine the mole fraction of each volatile component.

moles of methanol (M), CH3OH:	50.0 g		1 mol 32.04 g	=	1.56 mol
moles of water (W), H2O:	25.0 g		1 mol 18.01 g	=	1.39 mol

total number of moles = 1.56 + 1.39 = 2.95 mol

XM =

1.56 2.95

=

0.529

XD =

1.39 2.95

=

1 – 0.529 = 0.471

The partial pressure of each component:

P_M = X_MP° = (0.529)(93.3 torr) = 49.4 torr

P_W = X_WP° = (0.471)(17.5 torr) = 8.24 torr

The total pressure is the sum of the partial pressures:

P_T = P_M + P_W = 49.4 torr + 8.24 torr = 57.6 torr

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In distillation, the vapor is collected. The partial pressure of each component of the vapor is determined as above. Since the partial pressure of any substance is proportional to its number of moles, partial pressure can be used to determine the ratio of moles (mole fraction) within the vapor.

When a substance is distilled, it will boil at approximately the temperature of the component with the lowest boiling point. (Recall from Chapter 5 that the addition of a solute will raise the boiling point of the solvent. However, this is a small effect, so using the boiling point of the pure substance will not result in a significant error.) Also recall that the boiling point is the temperature when the vapor pressure equals the atmospheric pressure.

>> Example 3

If a mixture of 100.0 g of methanol (bp = 64.6 °C) and 25.0 g water (P° = 184 torr at 64.6°C) is distilled at 64.6 °C, what is the mole fraction of ethanol in the mixture?

Solution:

First, determine the partial pressure of both methanol and water in the vapor using Raoult's law. Raoult's law requires the mole fraction of each substance in the solution.

moles methanol =

100.0 g CH3OH

1 mol 32.04 g

=

3.12 mol

moles water =

25.00 g H2O

1 mol 18.01 g

=

1.39 mol

mol fraction methanol =

XM =

3.12 3.12 + 1.39

=

0.692

mol fraction water =

XW =

1.39 3.12 + 1.39

=

0.308

Since the distillation occurs at the boiling point, the vapor pressure of pure methanol is the atmospheric pressure. The boiling point in all tables assumes an atmospheric pressure of 1 atm or 760 torr.

partial pressure of methanol = P_M = X_MP° = (0.692)(760 torr) = 526 torr

partial pressure of water = P_W = X_WP° = (0.308)(184 torr) = 56.7 torr

total vapor pressure = 583 torr

The moles of methanol and moles of water in the vapor are proportional to the partial pressures. Therefore the ratio of partial pressures will also be equal to the ratio of moles.

XM (in vapor) =

526 torr 583 torr

=

0.903

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