# Raoult's Law

## Key Concepts

• The escaping tendency of a solvent is measured by its vapor pressure, which is dependent on temperature.
• Vapor pressure measures the concentration of solvent molecules in the gas phase.
• Assuming the solute is nonvolatile, the only particles in the gas phase are solvent molecules.
• In a solution, fewer solvent molecules are at the surface compared to the pure solvent, so a smaller proportion of solvent molecules will be in the gas phase and the vapor pressure for the solution is lower than that for the pure solvent.
Pa < Pao
where Pa = vapor pressure of the solution
and     Pao = vapor pressure of pure solvent
• Raoult's Law states that for an ideal solution the partial vapor pressure of a component in solution is equal to the mole fraction of that component times its vapor pressure when pure:
Pa = XaPao
where Pa = vapor pressure of the solution
and     Pao = vapor pressure of pure solvent
and     Xa = mole fraction of the solvent
• The fractional vapor pressure lowering is equal to the mole fraction of the solute:
Xb = (Pao - Pa) ÷ Pao
where Pa = vapor pressure of the solution
and     Pao = vapor pressure of pure solvent
and     Xb = mole fraction of the solute
• Fractional vapor pressure lowering can be used to calculate molecular mass (formula weight) or molar mass of a solute.

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## Example 1: Calculating the Vapor Pressure of a Solvent

(based on StoPGoPS approach to problem solving in chemistry)

Raoult's Law Problem:

1.00 g of nonvolatile sulfanilamide, C6H8O2N2S, is dissolved in 10.0 g of acetone, C3H6O.
The vapor pressure of pure acetone at the same temperature is 400 mm Hg.
Calculate the vapor pressure of the solution.
Solution to Raoult's Law Problem:

1. What is the question asking you to do?
Calculate the vapor pressure of the solution.

Pa = ? mm Hg

2. What information have you been given?
mass(solute) = mass(C6H8O2N2S(s)) = 1.00 g

mass(solvent) = mass(C3H6O) = 10.0 g

vapor pressure of pure solvent = Pao = 400 mm Hg

3. What is the relationship between what you know and what you need to find out?
Pa = XaPao

Pao = 400 mm Hg

Xa = mole fraction of solvent
= moles(solvent) ÷ (moles(solute) + moles(solvent))

4. Perform the calculations:

• Calculate moles of solute: n(C6H8O2N2S)
n(C6H8O2N2S) = mass(C6H8O2N2S) ÷ molar mass(C6H8O2N2S)
mass(C6H8O2N2S) = 1.00 g
molar mass(C6H8O2N2S) = (6 x 12.01) + (8 x 1.008) + (2 x 16.00) + (2 x 14.01) + 32.06 = 172.204 g mol-1

moles(C6H8O2N2S) = 1.00 g ÷ 172.204 = 0.005807 mol

• Calculate moles of solvent: n(C3H6O)
n(C3H6O) = mass(C3H6O) ÷ molar mass(C3H6O)
mass(C3H6O) = 10.0 g
molar mass(C3H6O) = (3 x 12.01) + (6 x 1.008) + 16.00 = 58.078 g mol-1

moles(C3H6O) = 10.0 g ÷ 58.078 = 0.1722 mol

• Calculate the mole fraction of the solvent: Xsolvent
Xsolvent = molessolvent ÷ (molessolute + molessolvent)
Xa = moles(C3H6O) ÷ [moles(C3H6O) + moles(C6H8O2N2S)]
= 0.1722 ÷ [0.1722 + 0.005807] = 0.9674
• Calculate the vapor pressure: Pa
Pa = XaPao
Pa = 0.9674 x 400 mm Hg = 386.96 mm Hg = 387 mm Hg
Addition of solute to solvent lowers the vapor pressure, that is,
Pa < Pao
where Pa is vapor pressure of solution and Pao is vapor pressure of pure solvent
Pao = 400 mm Hg (given in question)
Pa = 387 mm Hg (calculated)
387 < 400 so our answer is plausible.
6. State your solution to the problem.
Vapor pressure of solution is 387 mm Hg

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## Example 2: Calculating the Molecular Mass (Formula Weight) of a Solute

(based on StoPGoPS approach to problem solving in chemistry)

Raoult's Law Problem:

5.00 g of a nonvolatile compound was dissolved in 100 g of water at 30oC.
The vapor pressure of the solution was measured and found to be 31.20 Torr.
The vapor pressure of pure water at 30oC is 31.82 Torr.
Calculate the molar mass of the unknown solute.

Solution to Raoult's Law Problem:

1. What is the question asking you to do?
Calculate the molar mass of the solute.

molar masssolute = molar mass (b) = ? g mol-1

2. What information have you been given?
mass(solute) = m(b) = 5.00 g

mass(solvent) = m(a) = mass(H2O) = 100 g (at 30°C)

vapor pressure of solution = Pa = 31.20 Torr (at 30°C)

vapor pressure of pure solvent = Pao = 31.82 Torr (at 30°C)

3. What is the relationship between what you know and what you need to find?
(i) Pa = XaPao
Xa = Pa ÷ Pao

(ii) Xa = moles(a)/(moles(a) + moles(b))
Xamoles(a) + Xamoles(b) = moles(a)
Xamoles(b) = moles(a) - Xamoles(a)
moles(b) = (moles(a) - Xamoles(a))/Xa

Note: moles(a) = mass(H2O) ÷ molar mass(H2O)

(iii) moles(b) = mass(b) ÷ molar mass (b)
molar mass (b) = mass(b) ÷ moles (b)

4. Perform the calculations
• Calculate mole fraction of solvent: Xa
Xa = Pa ÷ Pao
Xa = 31.20 ÷ 31.82 = 0.9805
• Calculate moles of solute: moles(b)
moles(b) = (moles(a) - Xamoles(a))/Xa
moles(b) = ([100 g/(2 × 1.008 + 16.00) g mol-1] - 0.9805×[100 g/(2 × 1.008 + 16.00) g mol-1])/0.9805
moles(b) = ([100/18.016] - (0.9805×[100/18.016])/0.9805
moles(b) = (5.5066 - [0.9895 × 5.5066])/0.9805
moles(b) = 0.1074/0.9805 = 0.1095 mol = 0.110 mol
• Calculate molar mass solute: molar mass (b)
molar mass (b) = mass(b) ÷ moles (b)
molar mass (b) = 5.00 ÷ 0.110 = 45.45 g mol-1
Use an alternative method to calculate molar mass:
• Calculate the mole fraction of the solute: Xb
Xb = (Pao - Pa) ÷ Pao
Xb = (31.82 - 31.2) ÷ 31.82 = 0.0195
• Calculate moles of solvent: n(H2O)
n(H2O) = mass ÷ molar mass
n(H2O) = 100 g ÷ (2 x 1.008 + 16.00) g mol-1 = 5.5506 mol
• Calculate moles of solute: Xsolute
Xsolute = nsolute ÷ (nsolute + nsolvent)
0.0195 = nsolute ÷ (nsolute + 5.5506)
0.0195(nsolute + 5.5506) = nsolute
0.0195(nsolute) + 0.1082 = nsolute
0.1082 = nsolute - 0.0195(nsolute)
0.1082 = (1 - 0.0195)(nsolute)
0.1082 = 0.9805(nsolute)
(nsolute) = 0.1082 ÷ 0.9805 = 0.110 mol
• Calculate molar mass of solute: molar mass
molar mass = mass ÷ moles
molar mass = 5.00 g ÷ 0.110 mol = 45.45 g mol-1

This result is the same as we calculated above.
6. State your solution to the problem.
Molar mass of solute is 45.45 g mol-1

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