6.8 g `H_(2)O_(2)` is dissolved in 224 mL solution. This solution will be labelled as:

Options

A. `0.224V`
B. `20 V`
C. 5 V
D. 10 V

Explanation

To determine how to label the solution of 6.8 g of hydrogen peroxide (H₂O₂) dissolved in 224 mL of solution, we need to calculate the molarity of the solution. Here are the steps to solve the problem:

### Step 1: Calculate the number of moles of H₂O₂
To find the number of moles, we use the formula:

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

The molar mass of hydrogen peroxide (H₂O₂) can be calculated as follows:

- Hydrogen (H) = 1 g/mol × 2 = 2 g/mol
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol

Thus, the molar mass of H₂O₂ = 2 + 32 = 34 g/mol.

Now, we can calculate the number of moles:

\[ \text{Number of moles of H₂O₂} = \frac{6.8 \text{ g}}{34 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 2: Convert the volume of the solution from mL to L
Since molarity is expressed in moles per liter, we need to convert the volume from milliliters to liters:

\[ \text{Volume in L} = \frac{224 \text{ mL}}{1000} = 0.224 \text{ L} \]

### Step 3: Calculate the molarity of the solution
Molarity (M) is defined as the number of moles of solute divided by the volume of solution in liters:

\[ \text{Molarity (M)} = \frac{\text{Number of moles}}{\text{Volume in L}} = \frac{0.2 \text{ moles}}{0.224 \text{ L}} \approx 0.893 \text{ M} \]

### Step 4: Label the solution
The solution can be labeled based on its molarity. Since the calculated molarity is approximately 0.893 M, we can classify it as a concentrated solution of hydrogen peroxide. Solutions of hydrogen peroxide are often labeled as dilute or concentrated based on their molarity, with concentrations above 0.1 M typically considered concentrated.

### Final Answer:
The solution can be labeled as a **concentrated hydrogen peroxide solution**.

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