General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: solutions and colligative properties (202-SN1-RE)

This chapter links a measurable quantity, a temperature change or a pressure, to the NUMBER of dissolved particles. The whole difficulty comes down to two things: choosing the right concentration, molality or molarity depending on the formula, and never forgetting that electrolytes dissociate.

This sheet separates the five ways of expressing a concentration, fixes the four colligative laws with their units, and gathers the mistakes that repeat from one exam to the next, in particular on the factor $i$ and on the sign of the freezing-point depression. It is written for CEGEP students in General Chemistry, 202-SN1-RE (formerly 202-NYA), and it also prepares the colligative part of 202-SN2-RE.

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The thread of the chapter

A colligative property does not count grams or even moles of solute: it counts PARTICLES. That is why a mole of salt does twice the work of a mole of sugar, and why the factor $i$ decides everything.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

Five concentrations, and when each is used

• Molarity $C=\dfrac{n}{V_{solution}}$ in mol/L. It DEPENDS on temperature, since the volume expands. It is used for stoichiometry and osmosis.
• Molality $b=\dfrac{n}{m_{solvent}}$ in mol/kg, on the mass of the SOLVENT alone. Independent of temperature, it is used for freezing-point depression and boiling-point elevation.
• Mass percent $=\dfrac{m_{solute}}{m_{SOLUTION}}\times 100$ , on the total mass, never on that of the solvent.
• Mole fraction $x_{i}=\dfrac{n_{i}}{n_{total}}$ , without units, used by Raoult's law.
• Parts per million: for a dilute aqueous solution, $1$ ppm $=1$ mg/L. Dilution: $C_{1}V_{1}=C_{2}V_{2}$ .

Molarity relates to the volume of the SOLUTION, made up to the calibration mark; molality relates to the mass of the SOLVENT alone, weighed before the solute is added.

Molarity and molality are numerically close in DILUTE aqueous solution, since a litre of water weighs a kilogram. The gap becomes large as soon as the solution is concentrated, and that is precisely where exam exercises are set.

The four colligative laws

• Freezing-point depression: $\Delta T_{f}=i\,K_{f}\,b$ , with $K_{f}=1.86$ °C·kg/mol for water. The freezing temperature DECREASES: $T_{f}=0-\Delta T_{f}$ .
• Boiling-point elevation: $\Delta T_{b}=i\,K_{b}\,b$ , with $K_{b}=0.512$ °C·kg/mol. The boiling temperature INCREASES.
• Osmotic pressure: $\pi=i\,C\,R\,T$ , with $C$ in mol/L, $R=8.314$ kPa·L/(mol·K) and $T$ in kelvins.
• Raoult's law: $P_{solution}=x_{SOLVENT}\,P^{\circ}$ , so the lowering is $\Delta P=x_{solute}\,P^{\circ}$ .
• van't Hoff factor $i$ : number of particles released per formula unit. $1$ for a nonelectrolyte, $2$ for $NaCl$ , $3$ for $CaCl_{2}$ .

Same solution, same scale: the freezing point drops by 1.86 °C, the boiling point rises by only 0.51 °C. The two changes are in the ratio

K_{f}/K_{b}=3.6

MEASURED values of $i$ are always a little below the theoretical values, because of ion pairing in solution. The gap grows with concentration and with the charge of the ions.

Solubility: what makes it rise and what makes it drop

• Like dissolves like: an ionic or polar solute needs a polar solvent, a nonpolar solute a nonpolar solvent.
• For most SOLIDS, solubility INCREASES with temperature: this is what makes recrystallization possible.
• For GASES, solubility DECREASES with temperature, and increases with pressure: $S=k_{H}P$ , Henry's law.
• A saturated solution is at equilibrium with the undissolved solute: adding more solute no longer increases the concentration.
• Recrystallization: dissolve in the MINIMUM volume of boiling solvent, then cool slowly. You recover the starting mass minus what the cold solution still holds.

Two reference points, each in its own unit: the solubility of potassium nitrate climbs with temperature, that of dissolved oxygen drops. No single rule covers both cases.

The opposite behaviour of solids and gases explains two everyday facts: warm water holds less oxygen, which threatens fish in summer, and a lukewarm soft drink foams as soon as it is opened.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using molarity in a formula that requires molality

3 points, and an error that grows with concentration

What not to write

“The solution is $2.00$ mol/L, so $\Delta T_{f}=1.86\times 2.00=3.72$ °C.”

What to write

“Freezing-point depression uses MOLALITY, in moles per kilogram of SOLVENT. You must first convert from molarity to molality, which requires the density of the solution.”

Why: Colligative laws compare states at different temperatures, yet molarity changes with temperature since the volume expands. Only molality, defined on masses, stays constant.

2. Forgetting the van't Hoff factor for an electrolyte

the whole question, 5 points

What not to write

“ $1.00$ mol/kg of sodium chloride lowers the freezing point by $1.86$ °C.”

What to write

“ $NaCl$ releases TWO ions per formula unit, so $i=2$ and $\Delta T_{f}=2\times 1.86\times 1.00=3.72$ °C.”

At the SAME molality of

1.00

mol/kg, the depression doubles with salt and triples with calcium chloride. It is not the dissolved mass that counts, it is the number of particles released.

Why: A colligative property depends only on the NUMBER of dissolved particles, not on their nature. Dissolving a mole of salt amounts to dissolving two moles of independent particles.

3. Calculating molality on the mass of the solution

2 points, and a

14\%

error here

What not to write

“ $250$ g of solute in $1.50$ kg of water: the molality is $\dfrac{n}{1.75}$ .”

What to write

“Molality is calculated on the mass of the SOLVENT alone: $\dfrac{n}{1.50}$ . It is the mass percent that uses the mass of the solution.”

Molality divides by the mass of the SOLVENT alone, in green. The mass of the solution, in red, also contains the solute and gives a molality that is

14\%

too small.

Why: Molality measures how many particles there are for a fixed amount of solvent. Including the solute in it would make the quantity nonadditive and cost it its independence from temperature.

4. Adding the freezing-point depression instead of subtracting it

2 points, and a physical nonsense

What not to write

“ $\Delta T_{f}=4.99$ °C, so the solution freezes at $+4.99$ °C.”

What to write

“The freezing temperature DROPS: $T_{f}=0-4.99=-4.99$ °C. It is the boiling point that rises.”

The solute makes the freezing point DROP: from 0 °C, we subtract

\Delta T_{f}

. The red dot, above zero, is a lost sign.

Why: An antifreeze that made water freeze at $+5$ °C would be a coolant. The solute interferes with the organization of the crystal, so you must cool MORE to freeze.

5. Using the mole fraction of the solute in Raoult's law

3 points

What not to write

“ $x_{solute}=0.12$ and $P^{\circ}=3.17$ kPa, so $P_{solution}=0.12\times 3.17=0.38$ kPa.”

What to write

“ $P_{solution}=x_{SOLVENT}P^{\circ}=0.88\times 3.17=2.79$ kPa. The fraction of the solute gives the LOWERING, $0.38$ kPa, not the final pressure.”

The vapour pressure of the solution is the share of the SOLVENT,

88\%

P^{\circ}

. The share of the solute,

12\%

, is the lowering: mistaken for the final pressure, it gives the red bar.

Why: For a nonvolatile solute, only the solvent evaporates. The vapour pressure of the solution is therefore proportional to the proportion of solvent at the surface, not to that of the solute.

6. Forgetting kelvins or the factor $i$ in the osmotic pressure

the whole question, 5 points

What not to write

“ $\pi=CRT=0.150\times 8.314\times 37=46.1$ kPa for a sodium chloride solution at $37$ °C.”

What to write

“ $T=310$ K and $i=2$ : $\pi=2\times 0.150\times 8.314\times 310=773$ kPa.”

The solvent crosses the membrane toward the solution, whose level rises until the hydrostatic pressure compensates: it is this equilibrium pressure that

\pi

measures.

Why: The formula is modelled on that of ideal gases: it therefore requires kelvins. And like every colligative law, it counts particles, so it carries the factor $i$ .

7. Believing that the solubility of a gas increases with temperature

3 points on the reasoning question

What not to write

“As with sugar, hot water dissolves more carbon dioxide than cold water.”

What to write

“It is the OPPOSITE for gases: their solubility DECREASES as temperature rises. Warm water holds less oxygen, and a lukewarm soft drink goes flat when opened.”

When hot, the water holds LESS dissolved gas and the bubbles leave the beaker on the right: the opposite of sugar, which is more soluble when hot.

Why: Dissolving a gas releases energy, so heating shifts the equilibrium in the direction of release, in accordance with Le Chatelier's principle. For most solids, dissolution is on the contrary endothermic.

8. Taking the measured van't Hoff factor for the theoretical value

2 points on the commentary question

What not to write

“ $i$ is exactly $2$ for sodium chloride, whatever the concentration.”

What to write

“The MEASURED value is always a little lower, for example $1.87$ at $0.1$ mol/kg: some of the ions remain paired in solution.”

Two ion pairs among fifteen units count as only one particle each:

28

particles instead of

30

, so

i=1.87

and not

2

Why: Oppositely charged ions attract each other and form transient pairs that behave as a single particle. The gap grows with concentration and with the charge of the ions, which puts $CaCl_{2}$ further from its theoretical $i$ than $NaCl$ .

9. Dividing by the mass of the solvent in a mass percent

2 points

What not to write

“ $25$ g of solute in $100$ g of water make a $25\%$ solution by mass.”

What to write

“Mass percent relates to the mass of the SOLUTION: $\dfrac{25}{125}\times 100=20.0\%$ .”

Mass percent divides by the mass of the SOLUTION,

125

g, in green. Dividing by the

100

g of water, in red, gives

25\%

instead of

20.0\%

Why: Mass percent describes the composition of the liquid in the beaker, which weighs the sum of the two. Molality, for its part, makes the opposite choice and relates to the solvent alone.

10. Mixing units in a dilution

2 points, and a factor of

1000

What not to write

“ $C_{1}V_{1}=C_{2}V_{2}$ with $V_{1}=25$ mL and $V_{2}=0.500$ L.”

What to write

“Both volumes must be in the SAME unit: $V_{1}=0.025$ L and $V_{2}=0.500$ L, or else $25$ mL and $500$ mL.”

Why: The dilution relation is a ratio: it tolerates any volume unit, provided it is the same on both sides. It is the mixing that breaks it.

Which method to choose

Which concentration to use

The formula the concentration is going into

If freezing-point depression or boiling-point elevation → MOLALITY, in mol per kg of solvent

Example: $\Delta T_{f}=i\,K_{f}\,b$
If osmotic pressure → MOLARITY, in mol/L of solution

Example: $\pi=i\,C\,R\,T$
If Raoult's law or vapour pressure lowering → MOLE FRACTION, that of the solvent for the pressure, that of the solute for the lowering

Example: $P=x_{solvent}P^{\circ}$
If titration, dilution, stoichiometry in solution → MOLARITY, with $n=CV$ and $V$ in litres

Example: $C_{1}V_{1}=C_{2}V_{2}$
If the problem gives a density of the solution → it is the signal that you will have to convert from one concentration to another

without a density, you cannot convert a molarity into a molality

A density that is given and not used almost always signals a forgotten conversion between molarity and molality. It is the most reliable clue in the chapter.

Which factor $i$ to write

The nature of the solute

We count the particles that ONE formula unit releases: a neutral molecule stays whole, a salt separates into all its ions, a weak acid ionizes only in small part.

If neutral organic molecule: glucose, urea, glycerol, ethylene glycol, sucrose → $i=1$

Example: radiator antifreeze
If binary salt of the type $NaCl$ , $KBr$ , $MgSO_{4}$ → $i=2$

Example: road salt
If salt of the type $CaCl_{2}$ , $Na_{2}SO_{4}$ , $K_{2}CO_{3}$ → $i=3$

Example: calcium chloride, more effective per unit mass
If STRONG acid or STRONG base → complete dissociation, so $i$ equals the number of ions produced

Example: $HCl$ gives $i=2$
If WEAK acid or WEAK base → $i$ between $1$ and the number of ions, to be calculated with the degree of ionization

a weak acid ionized at $5\%$ gives $i=1.05$

Comparing two solutes at equal MASS and at equal MOLALITY does not give the same ranking: the molar mass comes into play in the first case, only the factor $i$ in the second. Read what the problem fixes.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a complete colligative problem

When to use it: The problem gives a mass of solute and a mass of solvent, and asks for a freezing temperature, a boiling temperature or a molar mass.

1 Convert the mass of solute to moles using its molar mass, writing the unit.
2 Calculate the molality by dividing by the mass of SOLVENT in kilograms, and say so explicitly.
3 Determine the factor $i$ , justifying it in one sentence: nonelectrolyte, binary salt, ternary salt.
4 Apply $\Delta T=i\,K\,b$ , then deduce the final temperature by SUBTRACTING for freezing and ADDING for boiling.
5 Conclude with the two temperatures and comment on how plausible the result is.

Concluding sentence

“The molality is $\dfrac{4.03}{1.50}=2.69$ mol/kg; ethylene glycol being a nonelectrolyte, $i=1$ , so $\Delta T_{f}=1.86\times 2.69=4.99$ °C and the solution freezes at $-4.99$ °C.”

The trap: Writing $\Delta T$ as the final answer. The question almost always asks for the TEMPERATURE, which still requires a subtraction or an addition, and this last step is worth a mark.

Marking: Typically 1 mark for the moles, 2 for the molality, 1 for the justified factor $i$ , 2 for the temperature change, 1 for the final temperature.

Determining a molar mass by osmometry

When to use it: The problem gives a mass of macromolecule, a volume of solution and a measured osmotic pressure.

1 Convert the temperature to kelvins and the pressure to a unit consistent with $R$ .
2 Isolate the concentration: $C=\dfrac{\pi}{iRT}$ , justifying that $i=1$ for a nonionic macromolecule.
3 Deduce the amount $n=CV$ , with $V$ in litres.
4 Calculate $M=\dfrac{m}{n}$ and check that the order of magnitude suits a macromolecule, that is, several tens of thousands of grams per mole.

Concluding sentence

“The concentration is $C=\dfrac{\pi}{RT}$ , hence $n=CV$ and $M=\dfrac{m}{n}$ , a molar mass of the order of $10^{4}$ g/mol, consistent with a protein.”

The trap: Using freezing-point depression for a macromolecule. The depression would be of the order of a thousandth of a degree, so unmeasurable: that is precisely why osmometry exists.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The direction of the temperature shift

Freezing must DROP below zero and boiling must RISE above one hundred. Any answer to the contrary is wrong with no further examination.

A solution that freezes at $+5$ °C: the sign of the depression has been lost.
The factor $i$ present in every colligative law

Go back over each formula and check that $i$ appears in it when the solute is an electrolyte. Its absence divides the result by two or three.

A depression of $1.86$ °C for a molality of $1$ mol/kg of salt: the factor $2$ is missing.
Boiling-point elevation about four times smaller than freezing-point depression

For water, $K_{f}=1.86$ and $K_{b}=0.512$ : the ratio is $3.6$ . The two calculated changes must respect this ratio.

A $\Delta T_{b}$ larger than the corresponding $\Delta T_{f}$ : the two constants have been swapped.
The order of magnitude of an osmotic pressure

A physiological solution develops several hundred kilopascals, that is, several atmospheres. A result of a few kilopascals signals degrees Celsius or a forgotten $i$ .

A $\pi=46$ kPa for physiological saline: the temperature was left in degrees Celsius.
The sum of the mole fractions

In a binary mixture, $x_{solvent}+x_{solute}$ must equal exactly $1$ , and the pressure lowering plus the final pressure must give back $P^{\circ}$ .

$0.88$ and $0.15$ : one of the two fractions is wrong.

The typical problem, taken apart

Radiator antifreeze, compared with road salt

$250$ g of ethylene glycol ( $M=62.07$ g/mol, nonelectrolyte) is dissolved in $1.50$ kg of water. Determine the molality of the solution, its freezing temperature and its boiling temperature.

What mass of sodium chloride ( $M=58.44$ g/mol) would have to be dissolved in the same mass of water to obtain the SAME freezing-point depression? Comment.

Data: $K_{f}=1.86$ °C·kg/mol, $K_{b}=0.512$ °C·kg/mol.

Two solutions in the same mass of water: we look for the mass of salt that gives the SAME depression as the

250

g of glycol.

Step 1

$n(\text{glycol})=\dfrac{250}{62.07}=4.03$ mol.

Why

We leave grams on the very first line, as in the whole previous chapter. Colligative laws only know amounts of substance.

Step 2

Molality: $b=\dfrac{4.03}{1.50}=2.69$ mol/kg, the mass of $1.50$ kg being that of the SOLVENT alone.

Why

Explicitly mentioning the solvent alone is worth a mark and avoids the error of dividing by $1.75$ kg, the mass of the complete solution.

Step 3

Ethylene glycol is a neutral molecule that does not dissociate: $i=1$ .

Why

The factor $i$ is justified, not asserted. Here the justification takes five words, but it separates a complete answer from one that guessed.

Step 4

$\Delta T_{f}=i\,K_{f}\,b=1\times 1.86\times 2.69=4.99$ °C, so $T_{f}=0-4.99=-4.99$ °C.

Why

The change is not the answer: the question asks for the temperature. Going from one to the other is a subtraction, and it is the exact place where answers stop too early.

Step 5

$\Delta T_{b}=1\times 0.512\times 2.69=1.38$ °C, so $T_{b}=100+1.38=101.38$ °C.

Why

Same molality, same factor $i$ , but a constant almost four times smaller and an opposite sign: a reminder that the two effects are not symmetrical.

Step 6

For the salt, $i=2$ , so the molality needed is $b=\dfrac{4.99}{2\times 1.86}=1.34$ mol/kg, that is $n=1.34\times 1.50=2.01$ mol and $m=2.01\times 58.44=118$ g.

Why

The factor $2$ halves the molality needed, and the lower molar mass reduces it again. Salt is therefore doubly more effective, per unit mass, than glycol.

Step 7

Comparison: $118$ g of salt does the work of $250$ g of glycol, half the mass for the same effect.

Why

This is the conclusion the question expects: the ranking of solutes follows neither the mass nor the molar mass alone, but the product of the factor $i$ and the number of moles.

Step 8

Comment: glycol is nevertheless used in radiators, because the ions of salt corrode the metals of the circuit, and salt is used on roads, where contact with the metal is brief.

Why

Colligative chemistry says what is most effective, it does not say what is usable. This nuance is almost always the last mark of the question.

Step 9

Check: $\dfrac{\Delta T_{f}}{\Delta T_{b}}=\dfrac{4.99}{1.38}=3.62$ , to compare with $\dfrac{K_{f}}{K_{b}}=\dfrac{1.86}{0.512}=3.63$ .

Why

This ratio depends neither on the molality nor on the solute: it therefore checks both calculations at once, without redoing any multiplication.

The conclusion, written out

“The glycol solution is $2.69$ mol/kg: it freezes at $-4.99$ °C and boils at $101.4$ °C. The same freezing-point depression would be obtained with only $118$ g of sodium chloride, half the mass, because salt releases two particles per formula unit.”

The classic mistake on this problem: Dividing by $1.75$ kg, the mass of the solution, instead of $1.50$ kg. We get $b=2.30$ mol/kg and a freezing point of $-4.28$ °C: the $0.7$ °C difference is small enough not to raise alarm and large enough to cost the whole question.

Learn by heart

• Molality $b=\dfrac{n}{m_{SOLVENT}}$ in mol/kg: freezing-point depression and boiling-point elevation. Molarity $C=\dfrac{n}{V_{SOLUTION}}$ : osmosis and titrations.
• Mass percent on the mass of the SOLUTION; molality on the mass of the SOLVENT.
• $\Delta T_{f}=i\,K_{f}\,b$ with $K_{f}=1.86$ ; the freezing point DROPS.
• $\Delta T_{b}=i\,K_{b}\,b$ with $K_{b}=0.512$ ; the boiling point RISES.
• $\pi=i\,C\,R\,T$ , with $T$ in KELVINS and $C$ in mol/L.
• Raoult: $P=x_{SOLVENT}P^{\circ}$ , and $\Delta P=x_{solute}P^{\circ}$ .
• $i=1$ nonelectrolyte, $2$ for $NaCl$ , $3$ for $CaCl_{2}$ ; measured values are a little lower.
• Solids: solubility that RISES with $T$ . Gases: solubility that FALLS with $T$ and rises with $P$ .

Frequently asked questions

What is the difference between molarity and molality?

Molarity is the number of moles of solute per litre of solution; it changes with temperature, since the volume expands. Molality is the number of moles per kilogram of solvent alone; it does not depend on temperature. That is why the freezing-point and boiling-point laws use molality.

Why does salt lower the freezing point twice as much as sugar?

Because a colligative property counts only the number of dissolved particles. A mole of sugar gives a mole of molecules, whereas a mole of sodium chloride dissociates into two moles of ions. The van't Hoff factor expresses exactly this, and it is one for sugar against two for salt.

Does the freezing point go up or down when a solute is added?

It goes down, always. The solute interferes with the organization of solvent molecules into a crystal, so you must cool more to freeze. The boiling point, for its part, goes up. The two effects have the same mechanism but opposite signs, and for water the first is about four times stronger than the second.

How do I choose between Raoult's law and freezing-point depression?

Look at the quantity asked for. If it is a vapour pressure, it is Raoult's law and you need a mole fraction. If it is a freezing or boiling temperature, the laws are freezing-point depression and boiling-point elevation and you need a molality. If it is an osmotic pressure, you need a molarity and kelvins.

Why does the solubility of a gas decrease when it is heated?

Because dissolving a gas releases energy. Heating therefore shifts the equilibrium in the direction of release, in accordance with Le Chatelier's principle. This is the opposite of most solids, whose dissolution absorbs energy. Warm water thus holds less oxygen, which threatens fish in summer.

Why is the measured van't Hoff factor lower than the theoretical value?

Because in solution, some of the oppositely charged ions remain paired and behave as a single particle. The effective number of independent particles is therefore a little smaller than predicted. The gap grows as the concentration increases and as the charges of the ions get larger.

Practise it

Corrected exercises: Solutions and colligative properties, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: solutions and colligative properties (202-SN1-RE)

The essentials

Five concentrations, and when each is used

The four colligative laws

Solubility: what makes it rise and what makes it drop

The mistakes that cost marks

1. Using molarity in a formula that requires molality

2. Forgetting the van't Hoff factor for an electrolyte

3. Calculating molality on the mass of the solution

4. Adding the freezing-point depression instead of subtracting it

5. Using the mole fraction of the solute in Raoult's law

6. Forgetting kelvins or the factor iii in the osmotic pressure

7. Believing that the solubility of a gas increases with temperature

8. Taking the measured van't Hoff factor for the theoretical value

9. Dividing by the mass of the solvent in a mass percent

10. Mixing units in a dilution

Which method to choose

Which concentration to use

Which factor iii to write

How the answer is expected to be written

Solving a complete colligative problem

Determining a molar mass by osmometry

Check before you hand in

The typical problem, taken apart

Radiator antifreeze, compared with road salt

Learn by heart

Frequently asked questions

What is the difference between molarity and molality?

Why does salt lower the freezing point twice as much as sugar?

Does the freezing point go up or down when a solute is added?

How do I choose between Raoult's law and freezing-point depression?

Why does the solubility of a gas decrease when it is heated?

Why is the measured van't Hoff factor lower than the theoretical value?

Corrected exercises: Solutions and colligative properties, 202-SN1-RE

See also

Looking for a 202-SN1-RE chemistry tutor in Montreal?

6. Forgetting kelvins or the factor $i$ in the osmotic pressure

Which factor $i$ to write