Volume Strength of Hydrogen Peroxide: Conversion to Molarity and % w/v

By Aniket Bhardwaj · 25 October 2026 · Updated 25 October 2026 · Class 11 Chemistry

Bottles of hydrogen peroxide in the lab or at the chemist are often labelled “20 volume” or “10 volume”. This is not a percentage and not a molarity. It is called the volume strength, and Class 11 and Class 12 papers ask you to convert it into molarity, grams per litre or per cent w/v. The conversion needs only one balanced equation and the molar volume of a gas.

A conversion flow shows 20 volume hydrogen peroxide becoming 1.786 M, 60.74 g/L, and 6.07% w/v. — 20 volume is about 6.07% w/v, not 20%.

What “volume strength” means

Hydrogen peroxide decomposes on heating or with a catalyst such as MnO₂:

2 H₂O₂ → 2 H₂O + O₂
2 mol (68 g) of H₂O₂ gives 1 mol of O₂ = 22.4 L at STP

A solution of “V volume” strength is one where 1 L of the solution releases V litres of O₂ at STP when all the H₂O₂ decomposes. So “20 volume H₂O₂” means 1 L of the solution gives 20 L of oxygen (or 1 mL gives 20 mL).

The conversion formulas

Molarity (M) = V ÷ 11.2
Strength in g/L = M × 34 (molar mass of H₂O₂ = 34.01 g/mol)
Per cent w/v = (g/L) ÷ 10

Why 11.2? From the equation, 1 mol of H₂O₂ gives half a mole of O₂, which is 22.4 ÷ 2 = 11.2 L at STP. So each mole of H₂O₂ in a litre supplies 11.2 L of gas, and the volume strength is 11.2 times the molarity. Always derive the number 11.2 from the equation in the exam; do not just recall it.

Which molar volume? The constant 22.4 L/mol applies at STP defined as 0°C and 1 atm. Some newer books use STP = 0°C and 1 bar, where the molar volume is 22.7 L/mol, and then the factor becomes 22.7 ÷ 2 = 11.35. Use the one your textbook or question states. This article uses 22.4, so the factor is 11.2.

Worked example 1: 20 volume H₂O₂ to molarity and per cent

Question. Find the molarity, strength in g/L and per cent w/v of 20 volume H₂O₂.

Step 1. Molarity = 20 ÷ 11.2 = 1.786 M.

Step 2. Strength = 1.786 × 34.01 = 60.74 g/L.

Step 3. Per cent w/v = 60.74 ÷ 10 = 6.07 g per 100 mL.

Check by direct route: 20 L of O₂ is 20 ÷ 22.4 = 0.8929 mol. Two moles of H₂O₂ are needed for each mole of O₂, so H₂O₂ = 2 × 0.8929 = 1.786 mol per litre. Same answer.

Answer. 1.79 M, about 60.7 g/L, about 6.07% w/v.

Worked example 2: from per cent w/v to volume strength

Question. A solution of H₂O₂ is 3.4% w/v. What is its volume strength?

Step 1. 3.4% w/v means 3.4 g in 100 mL, so 34 g in 1000 mL.

Step 2. Moles per litre = 34 ÷ 34.01 = 0.9997, about 1.00 M.

Step 3. Volume strength = 11.2 × 1.00 = 11.2 volume.

Answer. About 11.2 volume. Notice that 3.4% w/v is almost exactly 1 M, which is why this is a popular textbook choice.

Worked example 3: oxygen from a given sample

Question. How much oxygen at STP is released when 50 mL of 30 volume H₂O₂ decomposes completely?

Step 1. 30 volume means 1 mL of solution gives 30 mL of O₂.

Step 2. 50 mL × 30 = 1500 mL = 1.5 L of O₂.

Check: Molarity = 30 ÷ 11.2 = 2.679 M. Moles of H₂O₂ in 50 mL = 2.679 × 0.050 = 0.1339 mol. Moles of O₂ = 0.1339 ÷ 2 = 0.06696 mol. Volume = 0.06696 × 22.4 = 1.500 L. Correct.

Worked example 4: diluting 30 volume to 10 volume

Question. How much 30 volume H₂O₂ is needed to prepare 250 mL of 10 volume solution?

Step 1. Volume strength is proportional to concentration, so the dilution rule can be used directly: V₁S₁ = V₂S₂.

Step 2. V₁ × 30 = 250 × 10, so V₁ = 2500 ÷ 30 = 83.3 mL.

Step 3. Take 83.3 mL of the 30 volume solution and add water to make 250 mL.

Check in molarity: 30 volume = 2.679 M; 2.679 × 83.33 = 223.2 mmol. The 10 volume solution is 0.8929 M; 0.8929 × 250 = 223.2 mmol. The moles match.

Common mistakes

Using 22.4 instead of 11.2. The equation makes 1 mol of O₂ from 2 mol of H₂O₂. Dividing by 22.4 gives half the correct molarity.
Treating “20 volume” as 20%. 20 volume is about 6% w/v, not 20%.
Using the molar mass of water or of O₂. The molar mass for the g/L step is that of H₂O₂, 34 g/mol.
Forgetting the STP condition. Volume strength is defined at STP. Gas collected at room temperature needs a gas-law correction first.
Mixing 22.4 and 22.7 within one question.

Exam relevance

Given	Find	Method
Volume strength V	Molarity	M = V ÷ 11.2
Volume strength V	g/L and % w/v	M × 34, then ÷ 10
% w/v	Volume strength	g/L ÷ 34 gives M, then × 11.2
Volume of solution	O₂ released	Volume × V
Dilution	Volume to take	V₁S₁ = V₂S₂

This idea also appears in redox titrations of H₂O₂ with KMnO₄, where you must first know the molarity. Check which molar volume your syllabus uses before you start.

Convert a volume strength to molarity, then use the Molarity & Concentration tool to check your working.

Open the Molarity & Concentration Calculator →