Calculate the normality of 18 volume hydrogen peroxide solution .

Options

A. 2.31
B. 3.21
C. 5.6
D. 3.5

Answer

To calculate the normality of an 18 volume hydrogen peroxide (H2O2) solution, we can follow these steps:

### Step 1: Understand the meaning of "18 volume"
The term "18 volume" means that 1 liter of the hydrogen peroxide solution can liberate 18 liters of oxygen gas (O2) when it decomposes. This is important for determining how much H2O2 is present in the solution.

### Step 2: Write the decomposition reaction
The decomposition of hydrogen peroxide can be represented by the following balanced chemical equation:
\[ 2 H_2O_2 \rightarrow 2 H_2O + O_2 \]
From this equation, we see that 2 moles of H2O2 produce 1 mole of O2.

### Step 3: Calculate the amount of H2O2 needed for 18 liters of O2
We know that 1 mole of O2 occupies 22.4 liters at standard temperature and pressure (STP). Therefore, the number of moles of O2 produced from 18 liters is:
\[ \text{Moles of O}_2 = \frac{18 \text{ liters}}{22.4 \text{ liters/mole}} \approx 0.8036 \text{ moles} \]

From the balanced equation, we know that it takes 2 moles of H2O2 to produce 1 mole of O2. Thus, the moles of H2O2 needed for 0.8036 moles of O2 is:
\[ \text{Moles of H}_2O_2 = 2 \times 0.8036 \approx 1.6072 \text{ moles} \]

### Step 4: Calculate the mass of H2O2
The molar mass of H2O2 is approximately 34 g/mol. Therefore, the mass of H2O2 required is:
\[ \text{Mass of H}_2O_2 = 1.6072 \text{ moles} \times 34 \text{ g/mol} \approx 54.64 \text{ grams} \]

### Step 5: Calculate the strength of H2O2 in g/L
Since we have 54.64 grams of H2O2 in 1 liter of solution, the strength of the H2O2 solution is:
\[ \text{Strength} = 54.64 \text{ g/L} \]

### Step 6: Calculate the gram equivalent of H2O2
The equivalent weight of H2O2 is its molar mass divided by the number of electrons transferred in the reaction. In this case, the n-factor for H2O2 (for the decomposition reaction) is 1. Thus, the equivalent weight is:
\[ \text{Equivalent weight} = 34 \text{ g/mol} \]

Now, we can calculate the gram equivalents:
\[ \text{Gram equivalents} = \frac{\text{Strength}}{\text{Equivalent weight}} = \frac{54.64 \text{ g/L}}{34 \text{ g/equiv}} \approx 1.606 \text{ equivalents} \]

### Step 7: Calculate the normality
Normality (N) is defined as the number of gram equivalents of solute per liter of solution. Since we have calculated the gram equivalents:
\[ \text{Normality} = \frac{\text{Gram equivalents}}{1 \text{ L}} \approx 1.606 \text{ N} \]

### Final Answer
The normality of the 18 volume hydrogen peroxide solution is approximately **1.606 N**.