Colligative Properties: Vapour Pressure, Boiling Point and Freezing Point – Solvent Extraction and Colligative Properties (Senior 6 Chemistry) | Ganzaa

Solvent Extraction and Colligative Properties

Lessons

Senior 6 · Chemistry

Solvent Extraction and Colligative Properties

5 lessons

Lesson 4 of 5

Colligative Properties: Vapour Pressure, Boiling Point and Freezing Point

25 min

Does salt make water boil hotter?

Many cooks add salt to the cooking pot "so that the water boils hotter". A level teaspoon (about 6 g) of salt in 2 litres of water does raise the boiling point, but by about 0.05 °C, far too little to cook anything faster; you will calculate it yourself in this lesson. The same effect can be large, though. Sea water, with about 35 g of salt per kilogram, freezes at about −1.9 °C, and salt spread on frozen roads melts ice well below 0 °C. These effects depend only on how many solute particles are dissolved, not on what they are, and that makes them a way to count particles and so to find molar masses.

Definition

Colligative property: a property of a dilute solution that depends only on the number of solute particles in a given amount of solvent, not on their identity. The four are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure (Lesson 5).

Molality, $m$ : the amount of solute per kilogram of solvent, in $\mathrm{mol\,kg^{-1}}$ . Example: 0.100 mol of glucose in 500 g of water has $m = 0.200\,\mathrm{mol\,kg^{-1}}$ . Unlike concentration in $\mathrm{mol\,dm^{-3}}$ , molality does not change with temperature.

Ebullioscopic constant, $K_{b}$ , and cryoscopic constant, $K_{f}$ : the elevation of boiling point and the depression of freezing point caused by a solute at a molality of $1\,\mathrm{mol\,kg^{-1}}$ , for a given solvent, in $\mathrm{K\,kg\,mol^{-1}}$ . Example: for water $K_{b} = 0.512$ and $K_f = 1.86\,\mathrm{K\,kg\,mol^{-1}}$ .

van 't Hoff factor, $i$ : the number of particles actually present in solution for each formula unit dissolved. Example: $i \approx 2$ for NaCl in dilute solution (Na⁺ and Cl⁻), $i = 1$ for glucose.

Relative lowering of vapour pressure

Dissolve a non-volatile solute B, such as sugar, in a solvent A. Only solvent molecules can escape, and some of the surface is now occupied by solute particles, so by Raoult's law the vapour pressure falls to $p = x_A\,p_A^{\circ}$ . The lowering is $\Delta p = p_A^{\circ} - p = (1 - x_A)\,p_A^{\circ} = x_B\,p_A^{\circ}$ , so

$\frac{\Delta p}{p_A^{\circ}} = x_B = \frac{n_B}{n_A + n_B}$

The relative lowering of vapour pressure equals the mole fraction of the solute. It depends only on the number of solute particles, which is what makes it colligative. For a dilute solution $n_B \ll n_A$ , so $\frac{\Delta p}{p_A^{\circ}} \approx \frac{n_B}{n_A} = \frac{m_B M_A}{M_B m_A}$ , and the molar mass of the solute is

$M_B = \frac{m_B\,M_A\,p_A^{\circ}}{m_A\,\Delta p}$

where $m_{A}$ and $m_{B}$ are the masses of solvent and solute.

Worked example

Problem. 18.0 g of glucose, $\mathrm{C_6H_{12}O_6}$ ( $M = 180.2$ ), is dissolved in 180 g of water at 25 °C, where $p^{\circ}(\mathrm{H_2O}) = 3.17\,\mathrm{kPa}$ . Calculate the vapour pressure of the solution.

Solution.

$n(\text{glucose}) = \frac{18.0}{180.2} = 0.0999\,\mathrm{mol}$

$n(\mathrm{H_2O}) = \frac{180}{18.02} = 9.99\,\mathrm{mol}$

$x(\text{glucose}) = \frac{0.0999}{0.0999 + 9.99} = 0.00990 \qquad \Delta p = 0.00990 \times 3.17 = 0.0314\,\mathrm{kPa}$

Answer: $p = 3.17 - 0.0314 = 3.14\,\mathrm{kPa}$ . A 10 % sugar solution lowers the vapour pressure by only 1 %, which is why the method needs very precise pressure measurements.

Boiling point elevation and freezing point depression

A liquid boils when its vapour pressure equals the external pressure. A solution has a lower vapour pressure than the pure solvent at every temperature, so it must be heated further before its vapour pressure reaches 1 atm: the boiling point is raised. At the other end, a dilute solution freezes to give crystals of pure solvent. The solute particles lower the tendency of solvent molecules to leave the liquid for the solid, so the solution must be cooled further before crystals form: the freezing point is lowered (Figure 4.1).

For dilute solutions, both changes are proportional to the molality of solute particles:

$\Delta T_b = K_b\,m \qquad\qquad \Delta T_f = K_f\,m$

For an electrolyte, multiply by the van 't Hoff factor: $\Delta T_f = i\,K_f\,m$ . Because $m = \frac{m_B \div M_B}{m_A \div 1000}$ (with $m_{A}$ in grams), a measured $\Delta T$ gives the molar mass of the solute:

$M_B = \frac{K \times m_B \times 1000}{\Delta T \times m_A}$

Table 4.1: Constants for some solvents

Solvent	Boiling point / °C	$K_{b}$ / K kg mol⁻¹	Freezing point / °C	$K_{f}$ / K kg mol⁻¹
water	100.0	0.512	0.0	1.86
benzene	80.1	2.53	5.5	5.12
ethanoic acid	118	3.07	16.6	3.90
cyclohexane	80.7	2.79	6.5	20.0
camphor	–	–	179	37.7

$K_{f}$ is always larger than $K_{b}$ , and freezing points are easier to measure precisely, so freezing-point depression is the more common method. A Beckmann thermometer, which reads temperature differences to 0.01 K over a range of a few degrees, is used; the freezing point is taken from a cooling curve, allowing for supercooling. Solvents with a large $K_{f}$ , such as camphor (Rast's method) or cyclohexane, give large, easily measured depressions.

Figure 4.1 Phase diagram (sketch) of water (solid lines) and of an aqueous solution (dashed). The solute lowers the liquid–vapour curve, so the solution reaches 1 atm at a higher temperature ( $\Delta T_b$ ); the solid–liquid line moves to the left, so the solution freezes at a lower temperature ( $\Delta T_f$ ). The ice that forms is pure water, so the solid–vapour curve does not move.

Worked example

Problem. 2.40 g of a non-electrolyte Y is dissolved in 75.0 g of benzene. The freezing point falls by 0.960 K. Calculate the molar mass of Y. ( $K_{f}$ for benzene = 5.12 K kg mol⁻¹)

Solution.

$m = \frac{\Delta T_f}{K_f} = \frac{0.960}{5.12} = 0.1875\,\mathrm{mol\,kg^{-1}}$

$n(\mathrm{Y}) = 0.1875 \times 0.0750 = 0.01406\,\mathrm{mol}$

$M = \frac{2.40}{0.01406} = 171\,\mathrm{g\,mol^{-1}}$

Answer: $171\,\mathrm{g\,mol^{-1}}$ . (In one step: $M = \frac{5.12 \times 2.40 \times 1000}{0.960 \times 75.0} = 171$ .)

Worked example

Problem. (a) A solution of 0.100 mol of sodium chloride in 1.00 kg of water freezes at −0.348 °C. Calculate the van 't Hoff factor and the apparent degree of dissociation. (b) 1.20 g of ethanoic acid ( $M = 60.05$ ) in 50.0 g of benzene lowers the freezing point by 1.02 K. Find the apparent molar mass and explain it. ( $K_{f}$ : water 1.86, benzene 5.12 K kg mol⁻¹)

Solution. (a) Without dissociation, $\Delta T_f = 1.86 \times 0.100 = 0.186\,\mathrm{K}$ . Observed: 0.348 K.

$i = \frac{0.348}{0.186} = 1.87$

For $\mathrm{NaCl \rightarrow Na^+ + Cl^-}$ , $i = 1 + \alpha(2 - 1)$ , so $\alpha = 0.87$ . Attractions between the ions make them behave as if 87 % "dissociated"; at infinite dilution $i$ approaches 2.

(b) $M_{\text{app}} = \frac{5.12 \times 1.20 \times 1000}{1.02 \times 50.0} = 120\,\mathrm{g\,mol^{-1}}$

This is twice 60.05: in benzene, ethanoic acid molecules pair up through two hydrogen bonds into dimers, $\mathrm{(CH_3COOH)_2}$ , so there are half as many particles ( $i \approx 0.5$ ).

Warning

Common mistakes. (1) Using the mass of the solution, or mol dm⁻³, instead of molality per kilogram of solvent. (2) Forgetting to convert grams of solvent to kilograms. (3) Ignoring $i$ for ionic solutes: 0.1 mol kg⁻¹ $\mathrm{CaCl_2}$ lowers the freezing point about three times as much as 0.1 mol kg⁻¹ glucose. (4) Writing $\Delta T$ in °C and then adding 273: a temperature difference is the same in K and °C.

Tip

Count particles, not formula units. Every colligative formula has the same shape: change = constant × particles per kilogram. Glucose gives 1 particle, NaCl about 2, $\mathrm{CaCl_2}$ about 3, and an acid that dimerises about ½.

Khan Academy: Boiling point elevation and freezing point depression. Watch the molecular explanation of why dissolved particles raise the boiling point and lower the freezing point, and how the van 't Hoff factor counts ions.

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Practice

What is a colligative property? Name the four colligative properties.
Why is molality, rather than concentration in mol dm⁻³, used in $\Delta T = K m$ ?
12.0 g of a non-volatile solute X in 90.0 g of water has a vapour pressure of 3.100 kPa at 25 °C, where $p^{\circ}(\mathrm{H_2O}) = 3.169\,\mathrm{kPa}$ . Calculate the molar mass of X.
Find the boiling and freezing points of a solution of 3.42 g of sucrose ( $M = 342.3$ ) in 100 g of water.
What mass of urea, $\mathrm{CO(NH_2)_2}$ ( $M = 60.06$ ), must be dissolved in 250 g of water to lower its freezing point by 0.500 K?
0.500 g of a compound dissolved in 25.0 g of cyclohexane ( $K_{f} = 20.0$ ) lowers the freezing point by 2.35 K. Calculate its molar mass.
Put these 0.10 mol kg⁻¹ aqueous solutions in order of freezing point, highest first: $\mathrm{CaCl_2}$ , glucose, NaCl. Explain.
A car radiator holds a mixture of 300 g of ethane-1,2-diol ( $M = 62.07$ ) and 700 g of water. Calculate its freezing point, and evaluate why a colligative antifreeze must not be volatile.

Show answersHide answers

A property that depends only on the number of solute particles in a given amount of solvent, not on their nature: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure.
Molality is based on the mass of solvent, which does not change with temperature, whereas the volume of a solution (and so mol dm⁻³) does. It also counts particles per fixed amount of solvent, which is what the effect depends on.
$x_B = \frac{0.069}{3.169} = 0.0218$ ; $n(\mathrm{H_2O}) = 4.995\,\mathrm{mol}$ ; $n_B = \frac{0.0218 \times 4.995}{1 - 0.0218} = 0.111\,\mathrm{mol}$ ; $M = \frac{12.0}{0.111} = 108\,\mathrm{g\,mol^{-1}}$ (about $1.1 \times 10^{2}$ , since $\Delta p$ has only 2 significant figures).
$m = \frac{3.42 \div 342.3}{0.100} = 0.0999\,\mathrm{mol\,kg^{-1}}$ ; $\Delta T_b = 0.0511\,\mathrm{K}$ , boiling point 100.051 °C; $\Delta T_f = 0.186\,\mathrm{K}$ , freezing point −0.186 °C.
$m = \frac{0.500}{1.86} = 0.269\,\mathrm{mol\,kg^{-1}}$ ; $n = 0.269 \times 0.250 = 0.0672\,\mathrm{mol}$ ; mass $= 0.0672 \times 60.06 = 4.04\,\mathrm{g}$ .
$M = \frac{20.0 \times 0.500 \times 1000}{2.35 \times 25.0} = 170\,\mathrm{g\,mol^{-1}}$ .
Glucose (−0.19 °C) > NaCl (about −0.37 °C) > $\mathrm{CaCl_2}$ (about −0.56 °C). The depression is proportional to the number of particles: 1, about 2 and about 3 per formula unit.
$m = \frac{300 \div 62.07}{0.700} = 6.90\,\mathrm{mol\,kg^{-1}}$ ; $\Delta T_f = 1.86 \times 6.90 = 12.8\,\mathrm{K}$ , so the mixture freezes near −13 °C (the formula is only approximate at such a high molality). The solute must stay in the solution: a volatile antifreeze such as methanol would evaporate from a hot engine, and the protection would be lost.

Sources and credits

Constants in Table 4.1: standard values as tabulated in the CRC Handbook of Chemistry and Physics.
Image: Salting the road surface (49584798428).jpg by the Oregon Department of Transportation, via Wikimedia Commons, licensed under CC BY 2.0.
Video: Boiling point elevation and freezing point depression, Khan Academy, embedded from YouTube.
The text, phase diagram, table, worked examples and Practice items were written and drawn by Ganzaa.

1Does salt make water boil hotter?

2Relative lowering of vapour pressure

3Boiling point elevation and freezing point depression

Does salt make water boil hotter?

Relative lowering of vapour pressure

Boiling point elevation and freezing point depression