Step 1: Understand the definition of "x volume" of H\(_2\)O\(_2\).
"x volume" of H\(_2\)O\(_2\) means that 1 mL of the H\(_2\)O\(_2\) solution will produce 'x' mL of oxygen gas (O\(_2\)) at STP upon decomposition.
In this problem, we are given that 1 mL of the solution gives 20 mL of O\(_2\) at STP. Therefore, this is a "20 volume" H\(_2\)O\(_2\) solution.
Step 2: Relate volume strength to molarity.
The decomposition reaction is \( 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2 \).
This shows that 2 moles of H\(_2\)O\(_2\) produce 1 mole of O\(_2\).
1 mole of any gas at STP occupies 22400 mL.
So, 1 mole of O\(_2\) (22400 mL) is produced from 2 moles of H\(_2\)O\(_2\).
20 mL of O\(_2\) corresponds to \( \frac{20}{22400} \) moles of O\(_2\).
The moles of H\(_2\)O\(_2\) required to produce this are \( 2 \times \frac{20}{22400} = \frac{40}{22400} = \frac{1}{560} \) moles.
This amount of H\(_2\)O\(_2\) was present in 1 mL of the solution. So, the molarity (M) is:
Molarity \( M = \frac{\text{moles}}{\text{Volume (L)}} = \frac{1/560 \text{ moles}}{0.001 \text{ L}} = \frac{1000}{560} = \frac{25}{14} \) mol/L.
A useful shortcut is Volume Strength = 11.2 \(\times\) Molarity. So \( M = 20/11.2 \approx 1.78\). This confirms our calculation. (\(25/14 \approx 1.785\)).
Step 3: Convert molarity to (w/v) % strength.
(w/v) % means grams of solute per 100 mL of solution.
Strength (g/L) = Molarity (mol/L) \(\times\) Molar mass (g/mol).
Molar mass of H\(_2\)O\(_2\) is \(2(1) + 2(16) = 34\) g/mol.
Strength = \( \frac{25}{14} \times 34 = \frac{850}{14} = \frac{425}{7} \) g/L.
To find the percentage strength (grams per 100 mL), we divide the strength in g/L by 10.
(w/v) % = \( \frac{\text{Strength (g/L)}}{10} = \frac{425/7}{10} = \frac{42.5}{7} \).
\( \frac{42.5}{7} \approx 6.071 \)%.
This is approximately 6.06%.
