What is the concentration and % (w/w) of 5.6V `H2O2` solution? (Given molar mass of H2O2 = 34g/mol, density =1g/mL)

Options

A. 0.5,1.70
B. 0.25,1.70
C. 0.5,0.85
D. 0.25,0.85

Answer

To solve the problem of determining the concentration and % (w/w) of a 5.6 V H2O2 solution, we can follow these steps:

### Step 1: Understand Volume Strength
Volume strength (V) of a solution indicates how many mL of the solution can release 1 mole of the solute in terms of volume. For H2O2, a 5.6 V solution means that 1 L (1000 mL) of this solution can release 5.6 moles of O2.

### Step 2: Calculate Normality
The normality (N) of a solution can be calculated from its volume strength. The relationship is given by:
\[ N = \frac{V}{5.6} \]
Where V is the volume strength. Thus:
\[ N = \frac{5.6}{1} = 1 \, \text{N} \]

### Step 3: Determine the n-factor
The n-factor for H2O2 can be determined by considering the ions produced. H2O2 dissociates to give 2 H⁺ ions and 1 O₂²⁻ ion. Therefore, the total charge contributed by the ions is 2 (from H⁺) + 2 (from O₂²⁻) = 2. Hence, the n-factor is 2.

### Step 4: Calculate Molarity
Using the formula:
\[ \text{Molarity (M)} = \frac{\text{Normality (N)}}{\text{n-factor}} \]
Substituting the values:
\[ M = \frac{1}{2} = 0.5 \, \text{M} \]

### Step 5: Calculate % (w/w)
To find the % (w/w), we can use the formula:
\[ \text{Molarity (M)} = \frac{\text{Percentage (w/w)} \times \text{Density} \times 10}{\text{Molar Mass}} \]
Rearranging the formula to find % (w/w):
\[ \text{Percentage (w/w)} = \frac{\text{Molarity} \times \text{Molar Mass}}{\text{Density} \times 10} \]
Substituting the known values:
- Molarity = 0.5 M
- Molar Mass of H2O2 = 34 g/mol
- Density = 1 g/mL

Calculating:
\[ \text{Percentage (w/w)} = \frac{0.5 \times 34}{1 \times 10} = \frac{17}{10} = 1.7 \]

### Final Answers
- Concentration (Molarity) = 0.5 M
- % (w/w) = 1.7%