Concentration & Molarity — Drills · Grade 11 Science

Chapter: Solutions & Dissolved Particles

What these drills covers

By the end of Concentration & Molarity you should be able to:

6 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: concentration, molarity (M), mole, molar mass, dilution.

Problem 1

What is the molarity of a solution containing 4.0 mol of solute in 2.0 L of solution?
Answer and explanation

2 M

M = moles ÷ liters = 4.0 mol ÷ 2.0 L = 2.0 M.

Problem 2

Molarity is defined as…
  1. moles of solute per gram of solvent
  2. liters of solution per mole of solute
  3. grams of solute per liter of solution
  4. moles of solute per liter of solution
Answer and explanation

moles of solute per liter of solution

Molarity (M) equals moles of solute divided by liters of solution.

Problem 3

You dilute 60.0 mL of a 5.0 M solution with water to a final volume of 300.0 mL. What is the new molarity?
Answer and explanation

1 M

M₂ = M₁V₁ ÷ V₂ = (5.0 × 60.0) ÷ 300.0 = 300 ÷ 300 = 1.0 M.

Problem 4

What is the molarity of a solution made by dissolving 58.5 g of NaCl (molar mass 58.5 g/mol) in enough water to make 0.500 L of solution?
Answer and explanation

2 M

Moles = 58.5 ÷ 58.5 = 1.00 mol; M = 1.00 mol ÷ 0.500 L = 2.00 M.

Problem 5

When preparing a solution of an exact molarity, why do chemists say 'dissolve the solute and add water to make 1.0 L of solution' rather than 'add the solute to 1.0 L of water'?
Answer and explanation

Molarity depends on the final volume of the whole solution, not the volume of water you started with. Dissolving the solute changes the total volume slightly, so you must top up to the final line (often in a volumetric flask) to guarantee an exact 1.0 L of solution and therefore an exact molarity.

Problem 6

You want to dilute 40.0 mL of 6.0 M acid down to 1.5 M. What must the final volume be?
Answer and explanation

160 mL

V₂ = M₁V₁ ÷ M₂ = (6.0 × 40.0) ÷ 1.5 = 240 ÷ 1.5 = 160 mL.

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