The labeling on a bottle of `H_2 O_2` solution is 20 vol, then the concentration of `H_2 O_2` in percentage strength will be:

Options

A. 0.0303
B. 0.05
C. 0.0455
D. 0.0607

Answer

To find the percentage strength of a 20 volume `H2O2` solution, we can follow these steps:

### Step 1: Understand the meaning of "20 vol"
The term "20 volume" means that 1 liter of the `H2O2` solution can produce 20 liters of oxygen gas (`O2`) when it decomposes.

### Step 2: Write the decomposition reaction of `H2O2`
The decomposition reaction of hydrogen peroxide is:
\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]
From this reaction, we can see that 2 moles of `H2O2` produce 1 mole of `O2`.

### Step 3: Calculate the volume of `O2` produced from `H2O2`
From the definition of "20 volume":
- 1 liter of `H2O2` produces 20 liters of `O2`.

### Step 4: Use the ideal gas law to find the mass of `H2O2` needed
We know that 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP). Therefore, to find out how many grams of `H2O2` are needed to produce 20 liters of `O2`, we first need to find out how many moles of `O2` that corresponds to:
\[ \text{Moles of } O_2 = \frac{20 \text{ L}}{22.4 \text{ L/mol}} \approx 0.8929 \text{ moles} \]

### Step 5: Calculate the moles of `H2O2` required
From the reaction stoichiometry:
- 2 moles of `H2O2` produce 1 mole of `O2`.
Thus, to produce 0.8929 moles of `O2`, we need:
\[ \text{Moles of } H_2O_2 = 2 \times 0.8929 \approx 1.7858 \text{ moles} \]

### Step 6: Calculate the mass of `H2O2`
The molar mass of `H2O2` (Hydrogen Peroxide) is calculated as follows:
- Molar mass of `H2O2` = 2(Hydrogen) + 2(Oxygen) = 2(1) + 2(16) = 34 g/mol.

Now, we can calculate the mass of `H2O2` required:
\[ \text{Mass of } H_2O_2 = \text{Moles} \times \text{Molar Mass} = 1.7858 \text{ moles} \times 34 \text{ g/mol} \approx 60.67 \text{ g} \]

### Step 7: Calculate the percentage strength
The percentage strength is calculated using the formula:
\[ \text{Percentage Strength} = \left( \frac{\text{mass of solute (g)}}{\text{volume of solution (mL)}} \right) \times 100 \]
Since we have 60.67 g of `H2O2` in 1 liter (1000 mL) of solution:
\[ \text{Percentage Strength} = \left( \frac{60.67 \text{ g}}{1000 \text{ mL}} \right) \times 100 = 6.07\% \]

### Final Answer
The concentration of `H2O2` in percentage strength is approximately **6.07%**.

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