Concentrations of Solutions — Chemistry (Grade 11)

Define and explain Concentrations of Solutions in your own words
Use key terms such as concentration accurately
Apply what you have learned to new examples and questions
Avoid the common mistakes learners make with this topic

This lesson focuses on Concentrations of Solutions: work with moles per cubic decimetre.

Definition: Concentrations of Solutions

Work with moles per cubic decimetre.

Key ideas

The mole connects mass and particles

One mole of any substance contains 6.02 × 10²³ particles and has a mass in grams equal to its relative formula mass — so 18 g of water is one mole. The relationship moles = mass ÷ Mr lets you convert between the two in seconds. This is the key that unlocks every reacting-mass calculation.

Use the equation's molar ratio

A balanced equation gives the ratio of moles that react: in 2Mg + O₂ → 2MgO, 2 moles of Mg make 2 moles of MgO, a 1:1 ratio. The method is always the same: convert the known mass to moles, use the ratio to find moles of the unknown, then convert back to mass. If a reactant runs out first, it is the limiting reactant and it alone sets the maximum product.

Key term — concentration: Moles of solute per cubic decimetre of solution, in mol/dm³.

Worked example: Concentrations of Solutions

In the worked example, what mass of oxygen reacts with the 12.0 g of magnesium?

Moles of Mg = 0.494; moles of O₂ = 0.494 ÷ 2 = 0.247 mol; mass = 0.247 × 32.0 = 7.90 g. (Check: 12.0 + 7.90 = 19.9 g of product.).

Answer: Moles of Mg = 0.494; moles of O₂ = 0.494 ÷ 2 = 0.247 mol; mass = 0.247 × 32.0 = 7.90 g. (Check: 12.0 + 7.90 = 19.9 g of product.).

Common mistakes

Using the mass ratio instead of the mole ratio Grams do not react in the equation's ratio — moles do. Always convert masses to moles first, apply the ratio from the balanced equation, then convert back to grams.
Forgetting to double the oxygen in O₂ Mr of O₂ is 32, not 16 — the molecule has two atoms. The same trap applies to H₂, N₂ and the halogens: read the formula, not just the element.

Practice

5.84 g of NaCl is dissolved to make 1.00 dm³ of solution. What is its concentration? (Mr of NaCl = 58.4)

Convert grams to moles first.

Moles = 5.84 ÷ 58.4 = 0.100 mol; concentration = 0.100 ÷ 1.00 = 0.100 mol/dm³.

Calculate the number of moles in 10.0 g of sodium hydroxide, NaOH. (Ar: Na = 23.0, H = 1.0, O = 16.0)

First find Mr of NaOH.

Mr = 23.0 + 16.0 + 1.0 = 40.0. Moles = 10.0 ÷ 40.0 = 0.250 mol.

Calculate the concentration of a solution containing 0.20 mol of solute in 0.50 dm³ of solution.

Concentration = moles ÷ volume.

0.20 ÷ 0.50 = 0.40 mol/dm³.

A reaction should theoretically make 8.40 g of product but only 6.72 g is obtained. Calculate the percentage yield.

(actual ÷ theoretical) × 100.

(6.72 ÷ 8.40) × 100 = 80.0%.

Quick check

Concentrations of Solutions — quick check

Which of these best defines "concentration"?

Moles of solute per cubic decimetre of solution, in mol/dm³.

Explain why a chemist might prefer a reaction with 100% atom economy even if its percentage yield is lower.

High atom economy means most reactant atoms end up in the desired product rather than waste, saving raw materials and reducing waste disposal — even if some product is lost in practice.

Key takeaways

Concentrations of Solutions: work with moles per cubic decimetre.
The mole connects mass and particles: One mole of any substance contains 6.02 × 10²³ particles and has a mass in grams equal to its relative formula mass — so 18 g of water is one mole.
mole: The amount of substance containing 6.02 × 10²³ particles (Avogadro's constant).
Watch out for: using the mass ratio instead of the mole ratio

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