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Molarity practice problems

Molarity measures moles of solute per liter of solution. The denominator is the final solution volume, not just the water poured into the container.

By ExamCat · Original study lesson · Published September 24, 2026 · Updated September 25, 2026

Understand the method

Use M = n/V with V in liters. If the question gives grams, convert the mass to moles before calculating concentration. A useful check is dimensional: mol divided by L produces mol/L, whereas grams divided by milliliters does not.

Worked example, step by step

A student dissolves 5.85 g of NaCl and prepares 500 mL of solution. Using a molar mass of 58.5 g/mol, calculate molarity.

  1. Find the amount: 5.85 g ÷ 58.5 g/mol = 0.100 mol.
  2. Convert the final volume: 500 mL = 0.500 L.
  3. Divide amount by volume: 0.100 ÷ 0.500 = 0.200 mol/L.
  4. Interpret the result: each liter of a solution at this concentration contains 0.200 mol of NaCl formula units before considering dissociation.

A mistake to watch for

Do not multiply by two because NaCl separates into ions. The stated NaCl molarity describes the amount of dissolved NaCl; a question about total ideal particle concentration asks for something different.

Try it yourself: 12 questions

Write an answer before opening the explanation. For calculations, keep your units and intermediate steps; for concepts, state the reason for your choice.

1. How many moles are in 250 mL of 0.40 M solution?

Show answer and explanation

n = MV = 0.40 × 0.250 = 0.100 mol. Converting milliliters first prevents a thousandfold error.

2. What is the molarity of 0.30 mol in 1.5 L?

Show answer and explanation

0.30 ÷ 1.5 = 0.20 M. Doubling the volume at a fixed solute amount would halve this value.

3. Does adding 500 mL of water guarantee a final volume of 500 mL?

Show answer and explanation

No. The definition uses final solution volume. Prepare or measure that volume instead of assuming it equals the initial solvent volume.

4. Find the molarity of 0.075 mol of solute in 150 mL of solution.

Show answer and explanation

Convert 150 mL to 0.150 L, then M = 0.075/0.150 = 0.50 M. The amount is already in moles, so no molar-mass conversion is needed.

5. What volume of 0.25 M solution contains 0.050 mol of solute?

Show answer and explanation

V = n/M = 0.050/0.25 = 0.20 L, or 200 mL. Dividing mol by mol/L leaves liters.

6. How many grams of NaCl are needed for 250 mL of 0.20 M solution? Use 58.5 g/mol.

Show answer and explanation

n = MV = 0.20 × 0.250 = 0.050 mol. Mass = 0.050 × 58.5 = 2.925 g, about 2.9 g to two significant figures. The final solution volume must be 250 mL.

7. Find the molarity when 9.0 g of glucose makes 300 mL of solution. Use 180 g/mol.

Show answer and explanation

First n = 9.0/180 = 0.050 mol. Then M = 0.050/0.300 = 0.1667 M, about 0.17 M. Grams per liter alone would be a mass concentration, not molarity.

8. A 50.0 mL sample of 2.0 M stock is diluted to 200 mL. Find the new concentration.

Show answer and explanation

The sample contains 2.0 × 0.0500 = 0.10 mol. Divide by the final 0.200 L: 0.50 M. The amount stays constant while the volume increases fourfold.

9. Mix 100 mL of 0.20 M NaCl and 200 mL of 0.50 M NaCl. Assume volumes add and no solute is lost. Find the final concentration.

Show answer and explanation

Add moles, not concentrations: 0.100 × 0.20 + 0.200 × 0.50 = 0.120 mol. Total volume is 0.300 L, so M = 0.120/0.300 = 0.40 M. A simple average would incorrectly give equal weight to unequal volumes.

10. A solution contains 0.20 mol of nonvolatile solute in 0.50 L. Water evaporates until the volume is 0.40 L, with no solute lost. Find the new molarity.

Show answer and explanation

M = 0.20/0.40 = 0.50 M, up from 0.40 M. The amount of solute is unchanged; a smaller denominator raises concentration.

11. For an introductory ideal-dissociation problem, what is the chloride-ion concentration in 0.15 M CaCl₂?

Show answer and explanation

Each formula unit supplies two chloride ions, so [Cl⁻] = 2 × 0.15 = 0.30 M. Calcium-ion concentration is 0.15 M. Distinguish the stated solute molarity from the concentration of a particular ion.

12. A student calculates 0.040 mol ÷ 80 mL = 0.00050 M. Diagnose the error.

Show answer and explanation

The numerical result has units mol/mL, not mol/L. Convert 80 mL to 0.080 L first: 0.040/0.080 = 0.50 M. Missing the milliliter conversion creates a factor-of-1,000 error.

Build your next practice session

Mark the first step where your reasoning differed from the explanation. Turn that step into a question, answer it without looking, then change one condition in the example and explain what would change. A correct guess is a reason to revisit the method.

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Further reading

OpenStax Chemistry 2e: Molarity provides textbook background. The worked scenario and practice questions on this page were written independently for ExamCat.

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