Part 1 · Chapter 9

Solutions and Colligative Properties

How much is dissolved, how a solute changes vapour pressure, boiling and freezing points, and how counting particles weighs a molecule

Fundamentals of Chemistry Prof. Mithun Mondal Reading time ≈ 48 min
i What you'll learn
  • The kinds of solutions and how to express concentration: mole fraction, molarity, molality, ppm.
  • Henry's law for dissolved gases and Raoult's law for vapour pressure.
  • Ideal vs non-ideal solutions — positive and negative deviations and azeotropes.
  • The four colligative properties and why they depend only on the number of particles.
  • Using \(\Delta T_f,\ \Delta T_b,\ \pi\) to find a solute's molar mass.
  • Abnormal molar mass and the van't Hoff factor \(i\) for association and dissociation.
Section 9-1

Solutions & Their Types

A solution is a homogeneous mixture of two or more components. The component present in larger amount is the solvent; the others are solutes. Because any of the three states can dissolve in any other, there are nine classes — though liquid solutions are by far the most important.

SoluteSolventExample
GasLiquid\(\ce{O2}\) in water; soda water
LiquidLiquidethanol in water
SolidLiquidsalt or sugar in water
GasSolid\(\ce{H2}\) in palladium
SolidSolidalloys (brass, bronze)
Section 9-2

Expressing Concentration

Concentration can be quoted many ways; choose the one that stays constant under the conditions of the problem. Molarity changes with temperature (volume expands); molality and mole fraction do not, which is why colligative work uses them.

QuantityDefinitionTemperature-independent?
Mole fraction \(x\)\(x_A=\dfrac{n_A}{n_A+n_B}\)Yes
Molarity \(M\)\(\dfrac{\text{mol solute}}{\text{L solution}}\)No
Molality \(m\)\(\dfrac{\text{mol solute}}{\text{kg solvent}}\)Yes
Mass percent\(\dfrac{\text{mass solute}}{\text{mass solution}}\times100\)Yes
ppm\(\dfrac{\text{mass solute}}{\text{mass solution}}\times10^{6}\)Yes
Why molality for colligative properties? A boiling solution is hot and an osmosis experiment may not be — but the number of moles per kilogram of solvent never shifts with temperature, so molality keeps the arithmetic honest.
Section 9-3

Solubility & Henry's Law

The solubility of a solid in a liquid usually rises with temperature; the solubility of a gas falls with temperature but rises with pressure. Henry's law makes the pressure dependence quantitative: the partial pressure of a gas above a solution is proportional to its mole fraction in the solution.

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Henry's law
\(p = K_H\,x\)

Here \(p\) is the partial pressure of the gas, \(x\) its mole fraction in solution, and \(K_H\) the Henry constant. A larger \(K_H\) means a less soluble gas. This is why a sealed soda fizzes when opened — lowering \(p\) lowers \(x\).

Section 9-4

Vapour Pressure & Raoult's Law

Raoult's law states that for a volatile component the partial vapour pressure equals its mole fraction times the vapour pressure of the pure component. For a solution of two volatile liquids the total pressure is the sum of the two partials; for a non-volatile solute only the solvent contributes.

Raoult's law in two settings
\[ p_{\text{total}}=x_A p_A^\circ + x_B p_B^\circ \qquad\text{(two volatile liquids)} \]
\[ p_{\text{solution}}=x_{\text{solvent}}\,p^\circ \qquad\text{(non-volatile solute)} \]
Henry's law is Raoult's law with \(K_H\) in place of \(p^\circ\) — Raoult is the special case where the gas is the solvent itself.
pₐ p_B total p p xₐ=0 xₐ=1
An ideal solution: partials and total pressure are straight lines in mole fraction
Section 9-5

Ideal & Non-Ideal Solutions

An ideal solution obeys Raoult's law at every concentration, with \(\Delta H_{\text{mix}}=0\) and \(\Delta V_{\text{mix}}=0\) — the A–B forces match the A–A and B–B forces (e.g. benzene + toluene). Real mixtures often deviate.

BehaviourA–B forces\(\Delta H_{\text{mix}}\)AzeotropeExample
Ideal= A–A, B–B\(0\)nonebenzene + toluene
Positive deviationweaker\(>0\)minimum boilingethanol + water
Negative deviationstronger\(<0\)maximum boilingacetone + chloroform
Azeotropes resist distillation. A constant-boiling mixture vapourises in the same proportion it has in the liquid, so it cannot be separated by simple distillation — the reason absolute (100%) ethanol cannot be reached by distilling the ethanol–water azeotrope.
Section 9-6

Colligative Properties

A colligative property depends only on the number of solute particles, not on their nature. A mole of glucose and a mole of urea lower the freezing point of water equally; a mole of \(\ce{NaCl}\) lowers it about twice as much because it yields two ions. There are four classic colligative properties.

PropertyRelation
Relative lowering of vapour pressure\(\dfrac{p^\circ-p}{p^\circ}=x_{\text{solute}}\)
Elevation of boiling point\(\Delta T_b=K_b\,m\)
Depression of freezing point\(\Delta T_f=K_f\,m\)
Osmotic pressure\(\pi=MRT\)
Section 9-7

Relative Lowering of Vapour Pressure

A non-volatile solute occupies part of the surface, so fewer solvent molecules escape and the vapour pressure falls. Raoult's law gives the relative lowering directly as the mole fraction of solute — making it a route to molar mass.

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Relative lowering and molar mass
\(\dfrac{p^\circ-p}{p^\circ}=x_B=\dfrac{n_B}{n_A+n_B}\approx\dfrac{n_B}{n_A}\)

For a dilute solution \(\dfrac{p^\circ-p}{p^\circ}=\dfrac{w_B/M_B}{w_A/M_A}\), which rearranges to \(M_B=\dfrac{w_B\,M_A\,p^\circ}{w_A\,(p^\circ-p)}\).

Section 9-8

Boiling-Point Elevation & Freezing-Point Depression

Because the solute lowers the solvent's vapour pressure, the solution must be heated higher to boil and cooled lower to freeze. Both shifts are proportional to molality, with constants characteristic of the solvent.

The two thermal colligative properties
\[ \Delta T_b=K_b\,m,\qquad \Delta T_f=K_f\,m \]
\[ M_B=\dfrac{K_b\,w_B\times1000}{\Delta T_b\,w_A}=\dfrac{K_f\,w_B\times1000}{\Delta T_f\,w_A} \]
For water \(K_b=0.52\) and \(K_f=1.86\ \text{K kg mol}^{-1}\); benzene has \(K_f=5.12\). \(w_A\) is in grams.
Tf (pure) Tf (soln) ↓ Tb (pure) Tb (soln) ↑ solute widens the liquid range
The solute lowers freezing point and raises boiling point
Section 9-9

Osmosis & Osmotic Pressure

Across a semipermeable membrane — one that passes solvent but not solute — solvent flows from the dilute side to the concentrated side. The osmotic pressure \(\pi\) is the external pressure that must be applied to the solution to just stop this flow. It is the most sensitive colligative property, ideal for large molecules like proteins.

solvent solution semipermeable membrane level rises = π
Solvent crosses to the solution side until pressure π halts it
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van't Hoff equation for osmotic pressure
\(\pi=MRT=\dfrac{n_B}{V}RT \quad\Rightarrow\quad M_B=\dfrac{w_B RT}{\pi V}\)

Solutions of equal \(\pi\) are isotonic. The form \(\pi=MRT\) deliberately echoes the ideal-gas law — solute particles in a dilute solution behave much like a gas.

Section 9-10

Abnormal Molar Mass & the van't Hoff Factor

When a solute dissociates (like \(\ce{NaCl}\)) or associates (like benzoic acid dimerising in benzene), the number of particles differs from what the formula suggests, and the measured molar mass comes out "abnormal." The van't Hoff factor \(i\) corrects for this.

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The van't Hoff factor
\(i=\dfrac{\text{observed colligative effect}}{\text{normal (calculated) effect}}=\dfrac{\text{normal }M}{\text{observed }M}\)

Every colligative relation gains a factor \(i\): \(\Delta T_f=iK_f m\), \(\pi=iMRT\), and so on. For dissociation into \(n\) ions, \(i=1+(n-1)\alpha\); for association of \(n\) molecules, \(i=1-\left(1-\tfrac{1}{n}\right)\alpha\), where \(\alpha\) is the degree of dissociation or association.

Read the sign of \(i\). \(i>1\) means dissociation (more particles, larger effect, smaller apparent \(M\)); \(i<1\) means association (fewer particles); \(i=1\) means the solute stays as drawn.
Worked Examples

Putting It to Work

1 Molality and mole fraction

Problem. Find the molality and the mole fraction of glucose \((M=180)\) in a \(10\%\) (w/w) aqueous solution.

Solution. \(10\ \text{g}\) glucose sits in \(90\ \text{g}\) water:

Working
\[ m=\frac{10/180}{0.090}=0.62\ \text{mol kg}^{-1};\quad x=\frac{0.0556}{0.0556+5.0}=0.011 \]
2 Raoult's law, two liquids

Problem. Pure \(A\) and \(B\) have vapour pressures \(200\) and \(100\ \text{torr}\). Find the total pressure when \(x_A=0.40\).

Solution. Add the two partials:

Working
\[ p=0.40(200)+0.60(100)=80+60=140\ \text{torr} \]
3 Molar mass from VP lowering

Problem. Dissolving \(5.0\ \text{g}\) of a non-volatile solute in \(100\ \text{g}\) water drops the vapour pressure from \(23.8\) to \(23.4\ \text{torr}\). Find \(M_B\).

Solution. Relative lowering equals \(n_B/n_A\):

Working
\[ \frac{23.8-23.4}{23.8}=\frac{5.0/M_B}{100/18}\ \Rightarrow\ M_B\approx53.5\ \text{g mol}^{-1} \]
4 Molar mass from freezing point

Problem. \(1.00\ \text{g}\) of a non-electrolyte in \(50\ \text{g}\) benzene lowers the freezing point by \(0.40\ \text{K}\) \((K_f=5.12)\). Find \(M_B\).

Solution. Use the molar-mass form:

Working
\[ M_B=\frac{5.12\times1.00\times1000}{0.40\times50}=256\ \text{g mol}^{-1} \]
5 Molar mass from osmotic pressure

Problem. A solution of \(6.0\ \text{g}\) solute in \(1.0\ \text{L}\) shows \(\pi=2.0\ \text{atm}\) at \(300\ \text{K}\). Find \(M_B\) \((R=0.0821)\).

Solution. Rearrange \(\pi=MRT\):

Working
\[ M_B=\frac{w_B RT}{\pi V}=\frac{6.0\times0.0821\times300}{2.0\times1.0}\approx74\ \text{g mol}^{-1} \]
6 van't Hoff factor

Problem. A \(0.10\ m\) solution of \(\ce{NaCl}\) freezes at \(-0.35\ ^\circ\text{C}\) \((K_f=1.86)\). Find \(i\) and the degree of dissociation \(\alpha\).

Solution. Compare observed and normal \(\Delta T_f\); \(\ce{NaCl}\) gives \(n=2\):

Working
\[ i=\frac{0.35}{1.86\times0.10}=1.88;\quad i=1+(2-1)\alpha\ \Rightarrow\ \alpha=0.88 \]
Review

Chapter Summary

Concentration

Use molality and mole fraction for colligative work — they don't change with temperature.

Gas & vapour

Henry: \(p=K_H x\); Raoult: \(p=x_A p_A^\circ + x_B p_B^\circ\).

Deviations

Positive (weaker A–B, min-boiling azeotrope); negative (stronger A–B, max-boiling).

Colligative four

\(\tfrac{p^\circ-p}{p^\circ}=x_B\); \(\Delta T_b=K_b m\); \(\Delta T_f=K_f m\); \(\pi=MRT\).

Molar mass

Any colligative property yields \(M_B\); osmotic pressure is best for large molecules.

van't Hoff \(i\)

\(i>1\) dissociation, \(i<1\) association; insert \(i\) into every colligative relation.

Practice

Problems

Decide first whether the solute is volatile, non-volatile, or an electrolyte, then pick the matching relation. Take \(K_f=1.86\) and \(K_b=0.52\ \text{K kg mol}^{-1}\) for water unless told otherwise. Difficulty rises down the list.

  1. Calculate the molarity and molality of a \(98\%\) (w/w) \(\ce{H2SO4}\) solution of density \(1.84\ \text{g mL}^{-1}\).
  2. The Henry constant of \(\ce{CO2}\) in water is large. State, with reason, whether \(\ce{CO2}\) is more or less soluble than a gas with a small \(K_H\).
  3. Two liquids \(A\) and \(B\) have \(p_A^\circ=120\) and \(p_B^\circ=80\ \text{torr}\). Find the total vapour pressure of an equimolar mixture.
  4. Classify acetone + chloroform and ethanol + water as positive or negative deviations, and name the type of azeotrope each forms.
  5. The vapour pressure of water at \(25\ ^\circ\text{C}\) is \(23.8\ \text{torr}\). Find it after dissolving \(0.50\ \text{mol}\) urea in \(5.0\ \text{mol}\) water.
  6. \(2.0\ \text{g}\) of a solute in \(100\ \text{g}\) water raises the boiling point by \(0.052\ \text{K}\). Find its molar mass.
  7. A solution freezes at \(-1.86\ ^\circ\text{C}\). What is its molality, assuming a non-electrolyte?
  8. Calculate the osmotic pressure of \(0.01\ \text{M}\) glucose at \(300\ \text{K}\).
  9. A \(0.05\ m\) solution of \(\ce{K2SO4}\) is \(80\%\) dissociated. Find its van't Hoff factor and \(\Delta T_f\).
  10. Benzoic acid in benzene shows a molar mass of \(244\ \text{g mol}^{-1}\) instead of \(122\). Explain and find the degree of association.
  11. What concentration of \(\ce{NaCl}\) is isotonic with \(0.30\ \text{M}\) glucose? \((i_{\ce{NaCl}}=1.9)\)
  12. A \(5\%\) (w/v) solution of a protein has osmotic pressure \(0.018\ \text{atm}\) at \(300\ \text{K}\). Estimate its molar mass.
Tip: every colligative property is just a particle count in disguise. Convert the data to moles of particles, fold in the van't Hoff factor when the solute dissociates or associates, and the same four formulae carry you from vapour pressure to a molar mass.