Raoult's law

The pressure of a vapour

Pour some ethanol into a bottle and close the lid. Molecules at the surface escape into the space above. Some molecules of the vapour hit the surface and join the liquid again. After a short time, the two happen at the same rate. This is a dynamic equilibrium. The vapour now pushes on the walls with a fixed pressure. That is the vapour pressure of the liquid.

Three facts will help you:

• It depends on the liquid. Molecules that cling weakly escape easily, so that liquid has a high vapour pressure.
• It rises quickly with temperature, because more molecules have enough energy to escape.
• It does not depend on how much liquid there is, or on the size of the space above it. Only the liquid and the temperature matter.

A liquid boils when its vapour pressure becomes equal to the pressure outside. Here are four liquids at $298\ \text{K}$:

• Diethyl ether: $71.7\ \text{kPa}$. Its molecules do not form hydrogen bonds, so they leave easily.
• Methanol: $16.9\ \text{kPa}$.
• Ethanol: $7.87\ \text{kPa}$.
• Water: $3.17\ \text{kPa}$. Its molecules hold each other with strong hydrogen bonds.

We write the vapour pressure of a pure liquid as $p^\circ$, where the small circle means "pure". The unit is the kilopascal, $\text{kPa}$. Some books use mm Hg or bar. The law below is the same in every unit.

Two liquids mixed together

Now mix liquid A with liquid B. Both are volatile, so both escape. The vapour holds both of them. By Dalton's law, the pressure of the vapour is the sum of the partial pressures: $P = p_A + p_B$.

What fixes the partial pressure of A? Think of the surface of the mixture. Suppose 40 of every 100 surface molecules are A. Then A can leave from only 40 % of the places, so it escapes at 40 % of the rate of pure A. The French chemist Raoult found this rule in the 1880s.

Raoult's law. In a solution of volatile liquids, the partial pressure of each liquid is its mole fraction in the liquid times its vapour pressure when pure:

Recall the mole fraction, $x_A = \dfrac{n_A}{n_A + n_B}$, where $n$ is an amount in moles. The two mole fractions add up to $1$, so $x_B = 1 - x_A$.

Add the two partial pressures to get the total:

Put $x_B = 1 - x_A$ and tidy up:

This is the equation of a straight line. It runs from $p_B^\circ$ at $x_A = 0$, which is pure B, to $p_A^\circ$ at $x_A = 1$, which is pure A.

A worked example. Benzene (A) and toluene (B) at $298\ \text{K}$ have $p_A^\circ = 12.7\ \text{kPa}$ and $p_B^\circ = 3.79\ \text{kPa}$. The liquid has $x_A = 0.50$.

• $p_A = 0.50 \times 12.7\ \text{kPa} = 6.35\ \text{kPa}$
• $p_B = 0.50 \times 3.79\ \text{kPa} = 1.895\ \text{kPa}$
• $P = 6.35\ \text{kPa} + 1.895\ \text{kPa} = 8.245\ \text{kPa}$

The total lies between the two pure pressures, as the line in the picture says.

The vapour is not the same as the liquid

Dalton's law also gives the make-up of the vapour. The mole fraction of A in the vapour is $y_A$:

For the example, $y_A = \dfrac{6.35\ \text{kPa}}{8.245\ \text{kPa}} = 0.770$. The liquid is half benzene, but the vapour is 77 % benzene. Benzene has the higher vapour pressure, so it leaves faster. The vapour is richer in the more volatile liquid. Lesson 3 uses this fact to separate liquids by distillation.

Check that $y_A + y_B = 1$. Here $y_B = \dfrac{1.895\ \text{kPa}}{8.245\ \text{kPa}} = 0.230$, and $0.770 + 0.230 = 1$.

Set $x_A = 0$ and $x_A = 1$ and read the total pressure. Then set $x_A = 0.5$ with the pure pressures far apart. Which liquid dominates the vapour?

Practise with new numbers

A solid that does not evaporate

Now dissolve a solid in a liquid. Sugar in water is an example. The sugar does not evaporate, so the vapour above the solution holds water only. We call such a solute non-volatile.

Raoult's law still applies to the solvent. If $x_A$ is the mole fraction of the solvent, the vapour pressure of the solution is

The solute has mole fraction $x_B = 1 - x_A$, so $p = p^\circ(1 - x_B)$. Move $p^\circ$ and subtract:

The left side is the relative lowering of vapour pressure. It equals the mole fraction of the solute.

Why does the solute lower the pressure? Look at the surface.

Solute particles take up some of the places at the surface. Fewer solvent molecules are able to escape each second. The solvent also returns to the liquid at the usual rate, so the balance is reached at a lower pressure.

The lowering depends on how many solute particles there are, not on what they are. A property of this kind is called a colligative property.

A worked example. $68.4\ \text{g}$ of sucrose, of molar mass $342\ \text{g/mol}$, is dissolved in $360\ \text{g}$ of water at $298\ \text{K}$. The pure water has $p^\circ = 3.17\ \text{kPa}$.

• Sucrose: $n_B = \dfrac{68.4\ \text{g}}{342\ \text{g/mol}} = 0.200\ \text{mol}$.
• Water: $n_A = \dfrac{360\ \text{g}}{18.0\ \text{g/mol}} = 20.0\ \text{mol}$.
• $x_B = \dfrac{0.200\ \text{mol}}{20.0\ \text{mol} + 0.200\ \text{mol}} = 0.00990$.
• Lowering: $p^\circ - p = 0.00990 \times 3.17\ \text{kPa} = 0.0314\ \text{kPa}$.
• So $p = 3.17\ \text{kPa} - 0.0314\ \text{kPa} = 3.139\ \text{kPa}$.

The change is tiny. That is normal for a dilute solution, and it is why we must measure carefully.

Double the amount of solute. For a dilute solution the lowering nearly doubles. Now halve the mass of water. What happens to the lowering, and why?

Practise the lowering

Finding a molar mass

A measured lowering can give the molar mass of an unknown solute. Start with $\dfrac{p}{p^\circ} = \dfrac{n_A}{n_A + n_B}$ and turn it over:

This form is exact. Follow these steps:

1. Find $n_A$ from the mass of solvent and its molar mass.
2. Work out $\dfrac{p^\circ - p}{p}$ from the two pressures.
3. Multiply it by $n_A$ to get $n_B$.
4. Divide the mass of solute by $n_B$ to get its molar mass.

An example. $36.0\ \text{g}$ of a compound is dissolved in $100.0\ \text{g}$ of water at $298\ \text{K}$. The vapour pressure falls from $3.17\ \text{kPa}$ to $3.06\ \text{kPa}$.

• $n_A = \dfrac{100.0\ \text{g}}{18.0\ \text{g/mol}} = 5.556\ \text{mol}$.
• $\dfrac{p^\circ - p}{p} = \dfrac{3.17\ \text{kPa} - 3.06\ \text{kPa}}{3.06\ \text{kPa}} = 0.0359$.
• $n_B = 0.0359 \times 5.556\ \text{mol} = 0.200\ \text{mol}$.
• $M_B = \dfrac{36.0\ \text{g}}{0.200\ \text{mol}} = 180\ \text{g/mol}$.

The pressures differ in the third digit. A small error in either pressure gives a large error in $M_B$. This is a weakness of the method.

Raoult's law and Henry's law

A gas dissolved in a liquid also has a pressure above the liquid. For a dissolved substance B, Henry's law says $p_B = K_H\,x_B$, with a constant $K_H$ that belongs to the gas and the liquid. Raoult's law says $p_B = x_B\,p_B^\circ$. The two have the same form. Raoult's law is Henry's law in the special case $K_H = p_B^\circ$.

Try it in Python

The program prints the pressures above a mixture of benzene and toluene at every tenth of mole fraction.

Common mistakes

• Adding the pure vapour pressures. The total is $x_A p_A^\circ + x_B p_B^\circ$, which lies between the two.
• Using masses in the mole fraction. Divide each mass by its molar mass first.
• Taking the mole fraction in the vapour to be the one in the liquid. The vapour is richer in the more volatile liquid.
• Using the mole fraction of the solute in $p = x\,p^\circ$. It is the mole fraction of the solvent.
• Giving the lowering when the vapour pressure is asked. The lowering is $p^\circ - p$. The vapour pressure is $p$.

Remember these

• Vapour pressure: a fixed value for a liquid at one temperature. It does not depend on the amount of liquid.
• Raoult's law: $p_A = x_A p_A^\circ$. The total pressure of a mixture is $P = x_A p_A^\circ + x_B p_B^\circ$.
• The vapour has $y_A = \dfrac{p_A}{P}$. It is richer in the more volatile liquid.
• Non-volatile solute: $p = x_{solvent}\,p^\circ$ and $\dfrac{p^\circ - p}{p^\circ} = x_{solute}$.
• Molar mass from the lowering: $\dfrac{p^\circ - p}{p} = \dfrac{n_B}{n_A}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Liquid A has a vapour pressure of $40\ \text{kPa}$ when pure. In an ideal solution its mole fraction is $0.25$. What is its partial pressure?

2. Equal amounts of A and B are mixed. The vapour is 80 % A. Which liquid has the higher vapour pressure when pure?

3. Why is the vapour pressure of a sugar solution lower than that of pure water?

4. In a solution, 2 % of the molecules are solute molecules that cannot evaporate. By what fraction does the vapour pressure of the solvent fall?