`H_(2)O_(2)` solution used for hair bleaching is sold as a solution of approximately 5.0 g `H_(2)O_(2)` per 100 mL of the solution. The
molecular mass of `H_(2)O_(2)` is 34. The molarity of this solution is approximately

Options

A. `0.15 M`
B. `1.5 M`
C. `3.0 M`
D. `3.4 M`

Answer

To calculate the molarity of the hydrogen peroxide (H₂O₂) solution used for hair bleaching, we can follow these steps:

### Step 1: Calculate the number of moles of H₂O₂
The formula to calculate the number of moles is given by:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

Given:
- Mass of H₂O₂ = 5.0 g
- Molar mass of H₂O₂ = 34 g/mol

Now, substituting the values:

\[ \text{Number of moles} = \frac{5.0 \, \text{g}}{34 \, \text{g/mol}} \approx 0.147 \, \text{moles} \]

### Step 2: Convert the volume of the solution from mL to L
We know that:

\[ 1 \, \text{L} = 1000 \, \text{mL} \]

Given:
- Volume of solution = 100 mL

To convert to liters:

\[ \text{Volume in L} = \frac{100 \, \text{mL}}{1000} = 0.1 \, \text{L} \]

### Step 3: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute per liter of solution:

\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in L}} \]

Substituting the values we calculated:

\[ \text{Molarity (M)} = \frac{0.147 \, \text{moles}}{0.1 \, \text{L}} = 1.47 \, \text{M} \]

### Step 4: Round the molarity to an appropriate significant figure
Since the question asks for an approximate value, we can round 1.47 M to 1.5 M.

### Final Answer
The molarity of the H₂O₂ solution is approximately **1.5 M**.

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