Solutions Class 12 Chemistry: Colligative Properties Step by Step
Colligative properties is the part of Solutions where many Class 12 students lose marks, even after studying hard.
This post shows how to choose the right formula, when to apply the correction factor i, and how a parent can check progress at home.
What "colligative" means in plain words
A colligative property is a property of a solution that depends only on how many solute particles are present in a fixed amount of solvent. The kind of particle does not matter. The solute is what is dissolved. The solvent is what it dissolves in.
The syllabus has four such properties:
- Relative lowering of vapour pressure.
- Elevation of boiling point.
- Depression of freezing point.
- Osmotic pressure.
Each one has its own formula. A student who has memorised the formulas but cannot tell which property a question is asking about will pick the wrong one. This is the first pain point in the chapter.
Formula choice: match the question to the property
Ask the child to underline the "asked quantity" in the question before writing anything. The wording tells you the formula.
| If the question asks about | Use | Concentration unit needed |
|---|---|---|
| Vapour pressure of the solution | (p0 − p) / p0 = mole fraction of solute | Mole fraction |
| Boiling point rise | rise = boiling constant of solvent × m | Molality |
| Freezing point fall | fall = freezing constant of solvent × m | Molality |
| Osmotic pressure | π = C × R × T | Molarity |
Here m is molality, the moles of solute per kilogram of solvent. C is molarity, the moles of solute per litre of solution. Students often mix these two. Molality uses the mass of the solvent alone. Molarity uses the volume of the whole solution.
The unit column in the table is the most useful part. If the question gives grams of solute and grams of solvent, the child should think molality. If it gives the volume of solution, the child should think molarity.
Where the correction factor i comes in
The formulas above assume that the solute stays as one particle per formula unit. Glucose and urea do this. Salts such as NaCl do not. In water, NaCl splits into two ions, so it gives roughly twice the particles.
The factor i corrects for this. It is the observed colligative property divided by the value calculated without any correction. You can also write it as normal molar mass divided by observed molar mass.
With the correction, each formula gets one extra multiplier:
- Boiling point rise = i × boiling constant × m
- Freezing point fall = i × freezing constant × m
- π = i × C × R × T
- (p0 − p) / p0 = i × mole fraction of solute
The value of i depends on what the solute does in the solvent:
| Behaviour of solute | Effect on i | Example |
|---|---|---|
| Does not split or join | i = 1 | Glucose, urea |
| Splits into ions (dissociation) | i is greater than 1 | NaCl in water |
| Joins into larger units (association) | i is less than 1 | Ethanoic acid in benzene |
If a question says "complete dissociation", take i as the number of ions per formula unit. If it gives a degree of dissociation, α, then for a salt giving two ions, i = 1 + α. Read this wording carefully. It decides the answer.
A worked example, step by step
Question: 11.7 g of NaCl is dissolved in 500 g of water. Find the fall in freezing point. The freezing constant of water is 1.86 K kg mol−1. Assume complete dissociation. The molar mass of NaCl is 58.5 g mol−1.
- Pick the property. The question asks about freezing point, so use: fall = i × freezing constant × m.
- Find moles of solute. 11.7 / 58.5 = 0.2 mol.
- Find molality. The water is 500 g, which is 0.5 kg. So m = 0.2 / 0.5 = 0.4 mol per kg.
- Decide i. NaCl gives Na+ and Cl−, so i = 2.
- Calculate. Fall = 2 × 1.86 × 0.4 = 1.488 K.
The freezing point falls by about 1.49 K. Without i, the answer would be half of this, and the child would lose most of the marks.
Common mistakes to look for in the notebook
When you check your child's work, look for these slips. They are the usual reasons for lost marks.
- Grams used instead of kilograms. Molality needs the solvent in kg. This is the most common error.
- Molarity used in the freezing point formula. That formula needs molality.
- Forgetting i for salts. If the solute is an ionic compound, stop and ask what i is.
- Wrong direction for osmotic flow. Solvent moves through the semipermeable membrane from the dilute side to the concentrated side.
- Wrong sign for "elevation" and "depression". Boiling point goes up. Freezing point goes down. The formula gives only the size of the change. The child must add it to, or subtract it from, the pure solvent value.
- Stuck on molar mass questions. Many questions give a colligative property and ask for the unknown molar mass. Rearrange the formula and write the steps in order.
How a tutor helps with this chapter
This chapter is mostly about decision-making, not memory. A tutor can sit with the child and ask the same four questions on every problem. What property is asked? What unit does the formula need? Does the solute split or join? What does the final number mean?
In one-to-one teaching at home, the tutor can stop at the exact step where the child gets confused. In a large class this rarely happens. The child may be unsure whether to use molality or molarity, but the class moves on.
Great Home Tutors offers home tuition only. There is no coaching centre and no group batch. Families in Gurugram, including Sector 108 to 113, can ask for a chemistry home tutor by giving the class, subject, board and locality. A person calls back to understand the student. The first class is meant to show whether the fit is right before anything is fixed.
A weekly routine for parents in Gurugram
Class 12 students also have other subjects and school tests, so keep the routine short. Do not ask for hours of extra work. A simple plan is enough:
- Day 1: Read the NCERT text on the four properties. The child writes one line on each, in their own words.
- Day 2: Solve two NCERT problems that need no i. Write the unit of every quantity.
- Day 3: Solve two problems with salts. Write down the value of i before using any formula.
- Day 4: Solve two molar mass questions, working backwards from the colligative property.
- Day 5: Mix the types. The child first names the property and the unit, and only then calculates.
Ask your child to explain one problem to you aloud. You do not need to know chemistry. If they can say why they chose that formula, the idea has settled. If they cannot, go back to the first table.
Key takeaways
- Colligative properties depend on the number of dissolved particles, not their type.
- Choose the formula from what the question asks, then check whether it needs molality or molarity.
- Multiply by i when the solute splits into ions or joins into larger units.
- Convert the solvent mass to kilograms before finding molality.
- Short, regular practice with units written out works better than one long revision session.
Frequently asked questions
Is the factor i always a whole number?
No. For complete dissociation it matches the number of ions. When dissociation is partial, or when the solute associates, i can be a decimal. For a salt giving two ions, i = 1 + α.
Why does a salt lower the freezing point more than sugar of the same molality?
The salt splits into ions, so it gives more particles. Sugar stays as one particle per molecule. More particles mean a larger fall in freezing point.
Which formula uses molarity?
Osmotic pressure uses molarity. The boiling point and freezing point formulas use molality.
What is the difference between molality and molarity?
Molality is moles of solute per kilogram of solvent. Molarity is moles of solute per litre of solution. Molality does not change with temperature, because mass does not change. Molarity can change, because volume changes.
How much time should my child give this topic?
Follow a short, regular routine like the five-day plan above, and adjust it to the school schedule. Regular practice with units and the value of i matters more than the number of hours.
What to do next. Try it yourself on the page below.
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