Concentration & Molarity — Practice · Grade 11 Science

Chapter: Solutions & Dissolved Particles

What this practice covers

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

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

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

Problem 1

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

What is the molarity of a solution containing 3.0 mol of potassium chloride in 1.5 L of solution?
Answer and explanation

2 M

M = 3.0 mol ÷ 1.5 L = 2.0 M.

Problem 3

Describe how you would prepare 1.0 L of a 0.50 M NaCl solution. (The molar mass of NaCl is 58.5 g/mol.)
Answer and explanation

You need 0.50 mol/L × 1.0 L = 0.50 mol of NaCl, which is 0.50 mol × 58.5 g/mol ≈ 29.3 g. Weigh out about 29.3 g of NaCl, dissolve it in some water in a 1.0 L volumetric flask, then add water up to the 1.0 L line and mix.

Problem 4

You dilute 100.0 mL of a 2.0 M solution with water to a final volume of 500.0 mL. What is the new molarity?
Answer and explanation

0.4 M

M₂ = M₁V₁ ÷ V₂ = (2.0 × 100.0) ÷ 500.0 = 0.40 M.

Problem 5

What is the molarity of a solution made by dissolving 117 g of NaCl (molar mass 58.5 g/mol) in enough water to make 2.00 L of solution?
Answer and explanation

1 M

Moles = 117 ÷ 58.5 = 2.00 mol; M = 2.00 mol ÷ 2.00 L = 1.00 M.

Problem 6

Find molarity of 0.030 mol in 0.150 L solution. Show the reasoning, not just the conclusion.
Hints

Final solution volume is not necessarily the volume of added water.

Answer and explanation

M=0.030/0.150=0.20 mol/L.

Problem 7

A solution has 2 g solute in 100 g total solution. Find mass percent.
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

2/100×100%=2%; use solution mass, not solvent mass.

Problem 8

Dilution conserves solute amount only under the stated no-reaction conditionsDilution conserves solute amount only under the stated no-reaction conditionsBefore dilutionAfter dilutionc₁V₁ = nsolute = c₂V₂Same solute amount; larger solution volume lowers concentration.
Original equal-count dilution model; dot spacing and solvent volume are schematic rather than a molecular-scale simulation.
Interpret the supplied visual. Why does doubling final volume halve concentration in this selected dilution?
Answer and explanation

The unchanged solute amount is distributed through twice the solution volume, so n/V halves.

Dilution adds solvent while retaining the same amount of the specified solute, assuming no reaction, precipitation or loss. From n=cV, this gives c₁V₁=c₂V₂. The final volume V₂ is the total diluted solution volume, not merely the added solvent volume. The same expression should not be used automatically for reacting acid and base solutions, because the named solute amount can change. Draw a before-and-after entity inventory to identify what is actually conserved.

The unchanged solute amount is distributed through twice the solution volume, so n/V halves.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 9

Prepare 250. mL of 0.200 M solution by diluting a 1.00 M stock without reaction or loss. Find stock volume and explain whether 250. mL is added water volume.
Answer and explanation

V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.

V₁=c₂V₂/c₁=(0.200)(250.)/1.00=50.0 mL stock. Bring the mixture to 250. mL final solution volume; do not interpret that as adding exactly 250. mL water.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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