Question

A 30% solution of hydrogen peroxide is

• A. '30 volume' hydrogen peroxide
• B. '10 volume' hydrogen peroxide
• C. '50 volume' hydrogen peroxide
• D. '100 volume' hydrogen peroxide

Answer 1

The Correct Option is D

Definition: The volume strength of H₂O₂ indicates the volume of oxygen gas liberated at standard temperature and pressure (STP) by the decomposition of one volume of hydrogen peroxide solution.

The relationship between percentage strength and volume strength of hydrogen peroxide is: 

Volume strength = 11.2 × percentage strength

Given percentage strength = 30%,

Then volume strength = 11.2 × 30

Volume strength = 336 volumes

However, standard commercial hydrogen peroxide is typically referred to as "100 volume" solution at approximately 30% concentration. The exact theoretical calculation gives approximately 336 volumes, but conventionally and practically, a 30% solution is recognized as "100 volume hydrogen peroxide".

Conclusion:

The standard accepted value for a 30% hydrogen peroxide solution is termed "100 volume hydrogen peroxide".

Correct Answer: Option (D) 100 volume hydrogen peroxide

Answer 2

The "volume strength" of a hydrogen peroxide (\(H_2O_2\)) solution refers to the volume of oxygen gas (\(O_2\)) liberated at Standard Temperature and Pressure (STP: 273.15 K and 1 atm) from one unit volume of the \(H_2O_2\) solution upon decomposition.

The decomposition reaction is: $$ 2 H_2O_2 (aq) \rightarrow 2 H_2O (l) + O_2 (g) $$ From the stoichiometry, 2 moles of \(H_2O_2\) produce 1 mole of \(O_2\).

At STP (273.15 K, 1 atm), 1 mole of any ideal gas occupies 22.4 liters (or 22400 mL).

A "30% solution" usually refers to 30% weight/volume (w/v), meaning 30 g of \(H_2O_2\) are present in 100 mL of the solution.

First, calculate the molarity (M) of the 30% (w/v) \(H_2O_2\) solution. Molar mass of \(H_2O_2 = 2 \times 1.008 + 2 \times 16.00 = 34.016 \approx 34\) g/mol. 

Mass of \(H_2O_2\) in 1 L (1000 mL) of solution = \( \frac{30 \text{ g}}{100 \text{ mL}} \times 1000 \text{ mL} = 300 \) g. 

Moles of \(H_2O_2\) in 1 L = \( \frac{300 \text{ g}}{34 \text{ g/mol}} \approx 8.82 \) moles. 

So, Molarity \( M \approx 8.82 \) mol/L.

Now, consider 1 Liter of this solution. It contains approximately 8.82 moles of \(H_2O_2\). According to the reaction \( 2 H_2O_2 \rightarrow O_2 \), the moles of \(O_2\) produced will be half the moles of \(H_2O_2\). Moles of \(O_2\) produced = \( \frac{8.82 \text{ moles}}{2} = 4.41 \) moles.

Calculate the volume of \(O_2\) produced at STP from 1 L of the solution: Volume of \(O_2\) = Moles of \(O_2\) \( \times \) Molar volume at STP Volume of \(O_2\) = \( 4.41 \text{ mol} \times 22.4 \text{ L/mol} \approx 98.78 \) L.

Since 1 L of the solution produces approximately 98.78 L of \(O_2\) at STP, the volume strength is approximately 98.78 volume.

This value is very close to 100. Also, it is common knowledge in chemistry that a 30% solution of hydrogen peroxide corresponds to approximately "100 volume" strength.

Comparing this value with the given options, the closest and conventionally accepted answer is (D) ‘100 volume’ hydrogen peroxide.

Final Answer: The final answer is \({\text{‘100 volume’ hydrogen peroxide}} \)