Concentration Problems with Mixture Calculations
Concentration Problems deal with the amount of a substance present in a solution or mixture. Questions commonly involve finding concentration, quantity of solute, dilution, mixing solutions of different strengths, or replacing part of a mixture. Use percentage concentration, the alligation rule, and the replacement formula to solve these questions accurately.
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What Are Concentration Problems?
If a solution contains S units of solute and T units of total solution, then concentration = (S ÷ T) × 100%. The remaining quantity, T − S, is the solvent. For example, 20 litres of a 15% salt solution contains 0.15 × 20 = 3 litres of salt and 17 litres of water.
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Concentration Problems Formula & Tricks
Important Formulas
Use the total solution quantity as the denominator, not the quantity of solvent.
For a 12% solution of 25 litres, solute = (12 ÷ 100) × 25 = 3 litres.
In a 25-litre solution containing 3 litres of solute, the solvent quantity is 22 litres.
This gives the mixing ratio when two solutions produce a specified mean concentration.
When only solvent is added, the quantity of solute remains constant. C is concentration and V is volume.
Here n is the number of replacements, and the same quantity is removed and replaced each time.
Quick Tricks
When water or another pure solvent is added, solute quantity does not change. Equate the initial and final solute quantities instead of calculating every component.
The concentration after mixing two solutions must lie between the two original concentrations. If it does not, the data or calculation is incorrect.
Alligation uses the numerical differences between concentrations, not percentage ratios.
Concentration Problems Concepts
Finding Solute and Solvent Quantities
For a solution of T units and concentration p%, solute = (p/100)T. Solvent = T − (p/100)T, or ((100 − p)/100)T. The same method applies to litres, kilograms, grams, or any consistent unit.
Mixing Solutions by Alligation
If weaker concentration is a%, stronger concentration is b%, and the required concentration is m%, then weaker : stronger = (b − m):(m − a). The required concentration must lie between a% and b%.
Dilution by Adding Solvent
Apply C₁V₁ = C₂V₂ when the concentration and volume are expressed in compatible units. If the initial volume is V₁ and final volume is V₂, the amount of solvent added is V₂ − V₁.
Replacement of Part of a Solution
If V units are removed from a container of Q units, the fraction of the original solution remaining after one operation is (Q − V)/Q. After n identical operations, the remaining fraction is ((Q − V)/Q)ⁿ. This formula applies because the solution is mixed uniformly before each removal.
Concentration Problems Data Table
Use the structured data below while following the explanation and examples.
Concentration Calculation Examples
The table shows solute and solvent quantities for solutions with different concentrations.
| Total solution | Concentration | Solute quantity | Solvent quantity |
|---|---|---|---|
| 8 kg | 25% | 2 kg | 6 kg |
| 12 L | 15% | 1.8 L | 10.2 L |
| 20 L | 30% | 6 L | 14 L |
Note: Solute quantity = total solution × concentration percentage ÷ 100.
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Concentration Problems Revision Points
Use these formulas and rules for quick revision.
- Concentration = solute ÷ total solution × 100%.
- Solvent quantity = total solution − solute quantity.
- In dilution, C₁V₁ = C₂V₂ because solute quantity remains constant.
- For alligation, weaker : stronger = (stronger − mean):(mean − weaker).
- The final concentration after mixing must lie between the original concentrations.
- After n identical replacements, original solution left = initial quantity × (1 − removed/total)ⁿ.
- Keep units consistent before applying any formula.
Concentration Problems FAQs
How do you calculate the concentration of a solution?
Use Concentration = (quantity of solute ÷ total quantity of solution) × 100%. For 4 litres of solute in 20 litres of solution, concentration = (4/20) × 100 = 20%.
How much solute is present in 15 litres of a 20% solution?
Solute = (20/100) × 15 = 3 litres.
What is the concentration formula when water is added?
Use C₁V₁ = C₂V₂, since adding water does not change the quantity of solute. For example, 10 litres of 30% solution diluted to 15% gives final volume = (30 × 10)/15 = 20 litres.
How is the alligation ratio calculated?
For concentrations a%, b%, and mean m%, the ratio of weaker to stronger solution is (b − m):(m − a). For 10%, 40%, and a required 20%, the ratio is 20:10 = 2:1.
Can mixing two solutions produce a concentration higher than the stronger solution?
No. Without adding extra solute, the resulting concentration must lie between the concentrations of the two solutions.
What is the remaining original solution after repeated replacement?
If Q units are present and V units are replaced each time, the original solution left after n operations is Q × ((Q − V)/Q)ⁿ. For Q = 10, V = 2, and n = 2, it is 10 × (8/10)² = 6.4 units.