A sample of `H_(2)O_(2)` solution labelled as "28 volume" has density of 265 g/L. Mark the correct option(s) representing concentration of same solution in other units :
(a)`M_(H_(2)O_(2))=2.5`
(b)`%(w)/(V)=17`
(c)Mole fraction of `H_(2)O_(2)` = 0.2
(d)`m_(H_(2)O_(2))=13.88`

Options

A. `M_(H_(2)O_(2))=2.5`
B. `%(w)/(V)=17`
C. Mole fraction of `H_(2)O_(2)` = 0.2
D. `m_(H_(2)O_(2))=13.88`

Answer

To solve the problem, we need to find the concentration of hydrogen peroxide (H₂O₂) in different units based on the given information. Here are the steps to arrive at the solution:

### Step 1: Understand the Given Information
- The solution is labeled as "28 volume," which means it can produce 28 liters of oxygen (O₂) gas from 1 liter of the solution.
- The density of the solution is given as 265 g/L.

### Step 2: Calculate the Mass of H₂O₂ in the Solution
Since the density of the solution is 265 g/L, the mass of the solution (1 L) is:
\[ \text{Mass of solution} = 265 \text{ g} \]

### Step 3: Calculate the Number of Moles of O₂
Using the volume of O₂ produced:
\[ \text{Volume of O₂} = 28 \text{ L} \]
Using the molar volume of gas at STP (22.4 L/mol):
\[ \text{Moles of O₂} = \frac{28 \text{ L}}{22.4 \text{ L/mol}} = 1.25 \text{ moles} \]

### Step 4: Calculate the Number of Moles of H₂O₂
From the balanced reaction, 1 mole of H₂O₂ produces 0.5 moles of O₂. Therefore, the moles of H₂O₂ can be calculated as:
\[ \text{Moles of H₂O₂} = 2 \times \text{Moles of O₂} = 2 \times 1.25 = 2.5 \text{ moles} \]

### Step 5: Calculate Molarity (M) of H₂O₂
Molarity is defined as moles of solute per liter of solution:
\[ M_{H₂O₂} = \frac{\text{Moles of H₂O₂}}{\text{Volume of solution in L}} = \frac{2.5 \text{ moles}}{1 \text{ L}} = 2.5 \text{ M} \]

### Step 6: Calculate the Mass of H₂O₂
The molar mass of H₂O₂ is:
\[ \text{Molar mass of H₂O₂} = 2 \times 1 + 2 \times 16 = 34 \text{ g/mol} \]
Thus, the mass of H₂O₂ in the solution is:
\[ \text{Mass of H₂O₂} = \text{Moles of H₂O₂} \times \text{Molar mass of H₂O₂} = 2.5 \text{ moles} \times 34 \text{ g/mol} = 85 \text{ g} \]

### Step 7: Calculate the Mass of Water
The mass of water in the solution can be calculated as:
\[ \text{Mass of water} = \text{Total mass of solution} - \text{Mass of H₂O₂} = 265 \text{ g} - 85 \text{ g} = 180 \text{ g} \]

### Step 8: Calculate the Mole Fraction of H₂O₂
The number of moles of water (using its molar mass of 18 g/mol):
\[ \text{Moles of water} = \frac{180 \text{ g}}{18 \text{ g/mol}} = 10 \text{ moles} \]
Now, calculate the mole fraction of H₂O₂:
\[ \text{Mole fraction of H₂O₂} = \frac{\text{Moles of H₂O₂}}{\text{Moles of H₂O₂} + \text{Moles of water}} = \frac{2.5}{2.5 + 10} = \frac{2.5}{12.5} = 0.2 \]

### Step 9: Calculate the Mass Percent (w/v) of H₂O₂
Mass percent (w/v) is calculated as:
\[ \text{Mass percent (w/v)} = \frac{\text{Mass of H₂O₂}}{\text{Volume of solution in mL}} \times 100 = \frac{85 \text{ g}}{1000 \text{ mL}} \times 100 = 8.5\% \]

### Step 10: Calculate the Molality (m) of H₂O₂
Molality is defined as moles of solute per kg of solvent:
\[ \text{Mass of water in kg} = 180 \text{ g} = 0.18 \text{ kg} \]
\[ \text{Molality} = \frac{\text{Moles of H₂O₂}}{\text{Mass of water in kg}} = \frac{2.5}{0.18} \approx 13.88 \text{ mol/kg} \]

### Conclusion
Now we can summarize the results:
- Molarity \( M_{H₂O₂} = 2.5 \)
- Mass percent (w/v) = 8.5%
- Mole fraction of H₂O₂ = 0.2
- Molality \( m_{H₂O₂} = 13.88 \)

### Correct Options
From the calculations:
- (a) \( M_{H₂O₂} = 2.5 \) (Correct)
- (b) \( \% (w/v) = 17 \) (Incorrect, it is 8.5%)
- (c) Mole fraction of \( H₂O₂ = 0.2 \) (Correct)
- (d) \( m_{H₂O₂} = 13.88 \) (Correct)

### Final Answer
The correct options are (a), (c), and (d).