Boiling point and freezing point

A pinch of salt on an icy road

Sprinkle salt on an icy road and the ice melts, although the air is still below $0\ ^\circ\text{C}$. Pour antifreeze into a car radiator and the water stays liquid on a frosty night. It does not boil away on a hot day either.

Nobody changed the water. Something was dissolved in it. So a dissolved substance moves the temperature at which a liquid freezes, and the temperature at which it boils. How far does it move them?

Colligative properties

Some properties of a solution depend on how many particles are dissolved. They do not depend on what kind of particles they are. We call them colligative properties. The word means "depending on the collection".

There are four:

• the lowering of the vapour pressure of the solvent,
• the rise of the boiling point,
• the fall of the freezing point,
• the osmotic pressure.

We study the middle two here. Osmotic pressure comes next. Every result holds for a dilute solution of a solute that is not volatile, and that stays as single particles. A solute that splits into ions or joins into pairs needs one more idea, and we add it at the end.

Everything starts with the vapour pressure

A non-volatile solute lowers the vapour pressure of the solvent. The solute particles take up part of the surface, so fewer solvent molecules escape. Raoult's law gives the size of the effect:

Here $p_1^0$ is the vapour pressure of the pure solvent, $p_1$ is that of the solution, and $x_2$ is the mole fraction of the solute. The relative lowering equals the mole fraction of the solute. It depends only on the count of solute particles, so it is colligative.

The boiling point and the freezing point follow from this one fact.

The boiling point rises

A liquid boils when its vapour pressure becomes equal to the outside pressure. Look at the picture.

The blue curve is the pure solvent. It reaches the dashed line at the boiling point $T_b^0$. The red curve is the solution. It lies below the blue curve at every temperature, so at $T_b^0$ it has not reached the outside pressure. We have to heat the solution a little more. It boils at a higher temperature $T_b$.

The rise is $\Delta T_b = T_b - T_b^0$. For a dilute solution it is proportional to the molality $m$, the moles of solute in each kilogram of solvent:

The constant $K_b$ is the molal elevation constant. Its unit is $\text{K kg/mol}$. It belongs to the solvent and does not depend on the solute. It is the rise for $1\ \text{mol/kg}$, as if the solution stayed ideal.

The freezing point falls

A solid and a liquid are in balance when their vapour pressures are equal. For the pure solvent, the blue and grey curves cross at $T_f^0$. When we dissolve a solute, the solid that freezes out is still the pure solvent, so the grey curve does not move. The red curve is lower than the blue one, and it meets the grey curve further to the left. The solution freezes at a lower temperature $T_f$.

The fall is $\Delta T_f = T_f^0 - T_f$, and it is proportional to the molality:

The constant $K_f$ is the molal depression constant, also in $\text{K kg/mol}$. Here are some values:

• Water: $K_b = 0.512\ \text{K kg/mol}$ and $K_f = 1.86\ \text{K kg/mol}$.
• Benzene: $K_b = 2.53\ \text{K kg/mol}$ and $K_f = 5.12\ \text{K kg/mol}$.
• Ethanoic acid: $K_b = 3.07\ \text{K kg/mol}$ and $K_f = 3.90\ \text{K kg/mol}$.
• Camphor: $K_f = 39.7\ \text{K kg/mol}$.

Two things to notice. First, $K_f$ is bigger than $K_b$ for these solvents, so the freezing point moves more than the boiling point. Second, $\Delta T$ is a difference. A rise of $1\ \text{K}$ is the same as a rise of $1\ ^\circ\text{C}$. Never add $273$ to a difference.

Worked example: sweet water

A solution has $9.0\ \text{g}$ of glucose, of molar mass $180\ \text{g/mol}$, in $250\ \text{g}$ of water. Find its boiling point and its freezing point.

1. Moles. $\dfrac{9.0\ \text{g}}{180\ \text{g/mol}} = 0.050\ \text{mol}$.
2. Molality. $\dfrac{0.050\ \text{mol}}{0.250\ \text{kg}} = 0.20\ \text{mol/kg}$.
3. Boiling point. $\Delta T_b = 0.512\ \text{K kg/mol} \times 0.20\ \text{mol/kg} = 0.102\ \text{K}$, so the solution boils at $100.102\ ^\circ\text{C}$.
4. Freezing point. $\Delta T_f = 1.86\ \text{K kg/mol} \times 0.20\ \text{mol/kg} = 0.372\ \text{K}$, so it freezes at $-0.372\ ^\circ\text{C}$.

Look at the size of the two effects. The boiling point moves by a tenth of a degree, and the freezing point by more than three tenths. Both are small, so a good thermometer is needed.

Finding a molar mass

The fall of the freezing point gives the molality. The molality gives the moles. The moles and the mass give the molar mass. We can follow these steps one by one, or join them. For $w_2$ grams of solute in $w_1$ grams of solvent:

We want $M_2$, so we put $\Delta T = K\,m$ into this and solve:

Use $K_b$ with $\Delta T_b$, and $K_f$ with $\Delta T_f$. The masses are in grams, and the $1000$ changes grams of solvent into kilograms.

Worked example: sulphur in benzene. A solution of $3.84\ \text{g}$ of sulphur in $120\ \text{g}$ of benzene freezes $0.64\ \text{K}$ below pure benzene. Then, with $K_f = 5.12\ \text{K kg/mol}$, $w_2 = 3.84\ \text{g}$, $\Delta T_f = 0.64\ \text{K}$ and $w_1 = 120\ \text{g}$,

A sulphur atom has a mass of $32\ \text{g/mol}$, and $256\ \text{g/mol} \div 32\ \text{g/mol} = 8$. So a molecule of sulphur has eight atoms, $\text{S}_8$. The answer is a whole number of atoms, which tells us that the measurement makes sense.

Chemists often choose a solvent with a large $K_f$, such as camphor. A few milligrams of solute then give a fall that an ordinary thermometer can read.

Practise with new numbers

Try it in Python

The program turns a fall of the freezing point into a molar mass, in the three steps above.

Common mistakes

• Dividing by the mass of the solution. Molality uses the mass of the solvent only.
• Using grams of solvent in the molality. Change them to kilograms first.
• Adding $\Delta T_f$ to the freezing point. The freezing point goes down, so subtract.
• Using $K_f$ for a boiling point, or $K_b$ for a freezing point. Each constant has its own job.
• Changing a difference of temperature from $\text{K}$ to $^\circ\text{C}$ by adding $273$. A difference has the same number in both.

Remember these

• Colligative properties depend on the number of particles, not on their kind.
• $\Delta T_b = K_b\,m$ and $\Delta T_f = K_f\,m$, with $m$ in mol/kg and $K$ in K kg/mol.
• Water: $K_b = 0.512\ \text{K kg/mol}$ and $K_f = 1.86\ \text{K kg/mol}$.
• $M_2 = \dfrac{1000\,K\,w_2}{\Delta T\,w_1}$, with the masses in grams.

Check yourself

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

1. Why does a solution boil at a higher temperature than the pure solvent?

2. A solution is made of $0.30\ \text{mol}$ of a non-electrolyte in $600\ \text{g}$ of water. By how much does its freezing point fall?

3. Two solutions are made with the same mass of solvent. One has $0.10\ \text{mol}$ of glucose and the other has $0.10\ \text{mol}$ of a larger molecule that also does not split. Which one freezes lower?

4. The relative lowering of the vapour pressure of a solution is $0.020$. What is the mole fraction of the solute?