Class 12 Chemistry · Original EZT practice
Solutions (Class 12 Chapter 1): summary and practice
Solutions measures what is dissolved in what: molarity and molality for counting it, Raoult's law for how vapour pressure responds, and the four colligative properties - vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure - that depend only on how many particles are dissolved. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Expressing concentration
Molarity counts moles of solute per litre of SOLUTION; molality counts moles per kilogram of SOLVENT - so molality survives temperature changes while molarity shifts as the liquid expands. Mole fraction divides one component's moles by the total.
2. Solubility and Henry's law
Solubility is the maximum amount that dissolves at a given temperature and pressure. For gases in liquids, Henry's law says the dissolved amount is proportional to the gas's partial pressure - the fizz in a soda bottle is the classic case.
3. Vapour pressure and Raoult's law
Each volatile component contributes p = p0 x X, its pure vapour pressure times its mole fraction. A dissolved non-volatile solid lowers the solvent's vapour pressure in proportion to the solute's mole fraction.
4. Ideal and non-ideal solutions
Ideal solutions obey Raoult's law over the whole range, with no heat or volume change on mixing (like benzene with toluene). Non-ideal ones deviate positively (ethanol-water style, weaker new attractions) or negatively (stronger new attractions).
5. Colligative properties and abnormal molar mass
Colligative properties count particles, not their identity: relative lowering of vapour pressure, boiling-point elevation delta Tb = Kb m, freezing-point depression delta Tf = Kf m, and osmotic pressure pi = CRT. Section 1.7 adds the van't Hoff factor i for electrolytes that dissociate or associate, which rescales every colligative formula.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Compute molarity
Choose the method: M = moles of solute / litres of solution.
EZT practice question
4 g of NaOH (molar mass 40 g/mol) is dissolved to make 500 mL of solution. Find the molarity.
Show the worked solution
- Moles = 4/40 = 0.1 mol.
- Volume = 500 mL = 0.5 L.
- M = 0.1/0.5.
Answer: 0.2 mol/L (0.2 M).
Check: A full litre would hold 0.2 mol - here half a litre holds 0.1 mol, consistent.
Watch for this: Molarity uses the volume of the final SOLUTION, not the water you started with.
Now try it yourself
8 g of NaOH made up to 250 mL. Molarity?
Check your answer
0.8 M.
0.2 mol in 0.25 L gives 0.8 mol/L.
Compute molality
Choose the method: m = moles of solute / kg of solvent.
EZT practice question
18 g of glucose (180 g/mol) is dissolved in 500 g of water. Find the molality.
Show the worked solution
- Moles = 18/180 = 0.1 mol.
- Solvent = 500 g = 0.5 kg.
- m = 0.1/0.5.
Answer: 0.2 mol/kg (0.2 m).
Check: Molality ignores volume entirely - weigh the solvent, count the moles.
Watch for this: Do not divide by kg of SOLUTION - the denominator is solvent only.
Now try it yourself
10 g of glucose in 250 g of water. Molality?
Check your answer
0.222 m (approximately).
(10/180)/0.25 = 0.0556/0.25 = 0.222 m.
Find a mole fraction
Choose the method: X = moles of one component / total moles.
EZT practice question
46 g of ethanol (46 g/mol) is mixed with 90 g of water (18 g/mol). Find the mole fraction of ethanol.
Show the worked solution
- Ethanol: 46/46 = 1 mol. Water: 90/18 = 5 mol.
- X(ethanol) = 1/(1 + 5).
Answer: X(ethanol) = 1/6, about 0.167.
Check: X(water) = 5/6, and 1/6 + 5/6 = 1 - the two fractions must add to one.
Watch for this: Mole fraction has no units - if you wrote mol anywhere, redo the ratio.
Now try it yourself
23 g of ethanol in 90 g of water. Mole fraction of ethanol?
Check your answer
About 0.091.
0.5 mol ethanol vs 5 mol water: 0.5 ÷ 5.5 = 0.091.
Apply Raoult's law
Choose the method: p(component) = p0 x X; a non-volatile solute lowers the solvent's pressure.
EZT practice question
The vapour pressure of pure water at some temperature is 100 mmHg. A non-volatile solute brings the water's mole fraction to 0.8. Find the solution's vapour pressure and its lowering.
Show the worked solution
- p = p0 x X(water) = 100 x 0.8.
- Lowering = 100 - 80.
Answer: 80 mmHg, a lowering of 20 mmHg.
Check: Relative lowering equals the solute's mole fraction: 0.2 x 100 = 20 mmHg off.
Watch for this: The solute contributes nothing here - it is non-volatile, so only the solvent's term survives.
Now try it yourself
Same water, X(water) = 0.9. Vapour pressure?
Check your answer
90 mmHg.
100 x 0.9 = 90 mmHg.
Raise a boiling point
Choose the method: delta Tb = Kb x m.
EZT practice question
For water Kb = 0.52 K kg/mol. A solution has molality 0.5 m. Find the elevation of the boiling point.
Show the worked solution
- delta Tb = Kb x m.
- = 0.52 x 0.5.
Answer: delta Tb = 0.26 K, so the solution boils 0.26 degrees hotter than pure water.
Check: Halve the molality and the elevation halves - the relation is linear.
Watch for this: The answer is a temperature DIFFERENCE - the new boiling point is 100.26 degrees only at standard pressure.
Now try it yourself
Same water, 0.2 m solution. Elevation?
Check your answer
0.104 K.
0.52 x 0.2 = 0.104 K.
Measure osmotic pressure
Choose the method: pi = C R T, with C in mol/L and T in kelvin.
EZT practice question
Find the osmotic pressure of a 0.1 M solution at 300 K (R = 0.0821 L atm/K/mol).
Show the worked solution
- pi = C R T.
- = 0.1 x 0.0821 x 300.
Answer: pi = 2.46 atm (approximately).
Check: A 1 M solution would give 24.6 atm at the same temperature - ten times the concentration, ten times the pressure.
Watch for this: Convert Celsius to kelvin first - 27 degrees is 300 K, and using 27 directly ruins the answer.
Now try it yourself
A 0.05 M solution at 300 K. Osmotic pressure?
Check your answer
1.23 atm.
Half the concentration halves the pressure: 0.05 x 0.0821 x 300 = 1.23 atm.
Try a mixed set
Choose a method or explanation before opening each answer.
1. Molarity vs molality?
Show answer and reason
Molarity: moles per litre of solution; molality: moles per kg of solvent.
Molality is temperature-proof; molarity is not.
2. Raoult's law?
Show answer and reason
p = p0 x X for each volatile component.
Vapour pressure scales with mole fraction.
3. Boiling-point elevation?
Show answer and reason
delta Tb = Kb x m.
A colligative property - it counts particles.
4. Osmotic pressure?
Show answer and reason
pi = C R T.
T must be in kelvin.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- 10 g of glucose (180 g/mol) dissolved in 250 g of water. Find the molality.
- A 0.2 m aqueous solution with Kb = 0.52 K kg/mol. Find the boiling-point elevation.
- 8 g of NaOH (40 g/mol) made up to 250 mL. Find the molarity.
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Questions about this practice page
- What is the difference between molarity and molality?
- Molarity is moles of solute per litre of solution - it changes as temperature expands the liquid. Molality is moles per kilogram of solvent - mass does not change with temperature, so molality stays fixed.
- Why are they called colligative properties?
- Because they depend only on the NUMBER of dissolved particles, not on what the particles are: vapour-pressure lowering, boiling-point elevation, freezing-point depression and osmotic pressure all just count.
- What makes a solution ideal or non-ideal?
- An ideal solution obeys Raoult's law at every composition, with no heat or volume change on mixing. Non-ideal solutions deviate - positively when the new attractions are weaker than the pure ones, negatively when stronger.
- What is the van't Hoff factor?
- The factor i from section 1.7 that rescales colligative formulas when a solute dissociates (like NaCl, i near 2) or associates (like acetic acid dimers, i near 0.5). Forget it and electrolyte numericals come out wrong.
- Are these the NCERT exercise questions for Solutions (Class 12)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.