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Molarity of solutions: Class 11 practice

Molarity (M) is moles of solute per litre of solution. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Mole concept

The idea behind molarity of solutions

Molarity (M) is moles of solute per litre of solution. M = (mass ÷ molar mass) ÷ volume in litres. Dilution keeps moles constant: M₁V₁ = M₂V₂.

Step-by-step method

  1. Moles of solute = mass ÷ molar mass.
  2. Convert mL to L (÷1000).
  3. M = moles ÷ litres.
  4. For dilution use M₁V₁ = M₂V₂.

Worked example 1 (Basic)

8 g of NaOH is dissolved to make 250 mL of solution. Find its molarity.

  1. Moles = 8 ÷ 40 = 0.2
  2. Volume = 0.25 L
  3. M = 0.2 ÷ 0.25 = 0.8 M
  4. Answer: 0.8 M

Worked example 2 (Standard)

45 g of glucose (C₆H₁₂O₆) is dissolved to make 1000 mL of solution. Find its molarity.

  1. Moles = 45 ÷ 180 = 0.25
  2. Volume = 1 L
  3. M = 0.25 ÷ 1 = 0.25 M
  4. Answer: 0.25 M

Worked example 3 (Challenge)

500 mL of 0.4 M KOH solution is diluted to 2500 mL. Find the new molarity.

  1. M₁V₁ = M₂V₂
  2. M₂ = 0.4 × 500 ÷ 2500 = 0.08
  3. Answer: 0.08 M

Common mistakes and misconceptions

  • Dividing by mL instead of L
  • Using volume of solvent instead of solution

Practice questions

  1. (Basic) 11.700000000000001 g of NaCl is dissolved to make 1000 mL of solution. Find its molarity.
  2. (Basic) 20 g of NaOH is dissolved to make 250 mL of solution. Find its molarity.
  3. (Standard) 90 g of glucose (C₆H₁₂O₆) is dissolved to make 250 mL of solution. Find its molarity.
  4. (Standard) 20 g of NaOH is dissolved to make 1000 mL of solution. Find its molarity.
  5. (Challenge) 500 mL of 0.5 M glucose (C₆H₁₂O₆) solution is diluted to 1000 mL. Find the new molarity.
  6. (Challenge) 250 mL of 0.8 M NaCl solution is diluted to 1000 mL. Find the new molarity.

Answer key with full working

  • Q1: Moles = 11.700000000000001 ÷ 58.5 = 0.2 → Volume = 1 L → M = 0.2 ÷ 1 = 0.2 M → Answer: 0.2 M
  • Q2: Moles = 20 ÷ 40 = 0.5 → Volume = 0.25 L → M = 0.5 ÷ 0.25 = 2 M → Answer: 2 M
  • Q3: Moles = 90 ÷ 180 = 0.5 → Volume = 0.25 L → M = 0.5 ÷ 0.25 = 2 M → Answer: 2 M
  • Q4: Moles = 20 ÷ 40 = 0.5 → Volume = 1 L → M = 0.5 ÷ 1 = 0.5 M → Answer: 0.5 M
  • Q5: M₁V₁ = M₂V₂ → M₂ = 0.5 × 500 ÷ 1000 = 0.25 → Answer: 0.25 M
  • Q6: M₁V₁ = M₂V₂ → M₂ = 0.8 × 250 ÷ 1000 = 0.2 → Answer: 0.2 M

Syllabus note

This skill is listed under "Some Basic Concepts of Chemistry" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is molarity of solutions in?

"Some Basic Concepts of Chemistry", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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