A sample of `H_(2)O_(2)` solution labelled as 56.75 volume has density of 530 gm/L. Mark the correct options (s) representing concentration of same solution in other units. (Solution contains only `H_(2)O and H_(2)O_(2)`)

Options

A. `M_(H_(2)O_(2)) = 6`
B. `% (w)/(v) = 17`
C. Mole fraction of `H_(2)O_(2) = 0.25`
D. `m_(H_(2)O_(2)) = (1000)/(72)`

Answer

To solve the problem, we need to determine the concentration of the hydrogen peroxide (H₂O₂) solution in various units based on the given volume strength and density. Here’s a step-by-step breakdown of the solution:

### Step 1: Understanding Volume Strength
The volume strength of a solution indicates how many milliliters of oxygen gas (O₂) are produced by one liter of the solution at standard temperature and pressure (STP). For H₂O₂, a volume strength of 56.75 means that 1 liter of this solution will produce 56.75 liters of O₂.

### Step 2: Relating Volume Strength to Molarity
The relationship between volume strength (Vs) and molarity (M) for H₂O₂ is given by the formula:
\[ Vs = 11.2 \times M \]
Given that the volume strength is 56.75, we can set up the equation:
\[ 56.75 = 11.2 \times M \]

### Step 3: Calculating Molarity
To find the molarity (M), we rearrange the equation:
\[ M = \frac{56.75}{11.2} \]
Calculating this gives:
\[ M \approx 5.06 \text{ mol/L} \]
Rounding this, we can say:
\[ M \approx 5 \text{ mol/L} \]

### Step 4: Calculating Mass of H₂O₂
Using the molarity, we can calculate the number of moles of H₂O₂ in 1 liter of solution:
\[ \text{Moles of H₂O₂} = M \times \text{Volume (L)} = 5 \text{ mol/L} \times 1 \text{ L} = 5 \text{ moles} \]

Next, we calculate the mass of H₂O₂ using its molar mass (34 g/mol):
\[ \text{Mass of H₂O₂} = \text{Moles} \times \text{Molar Mass} = 5 \text{ moles} \times 34 \text{ g/mol} = 170 \text{ g} \]

### Step 5: Finding Mass of the Solution
The density of the solution is given as 530 g/L. Thus, the mass of 1 liter of the solution is:
\[ \text{Mass of solution} = 530 \text{ g} \]

### Step 6: Calculating Mass of Solvent (Water)
The mass of the solvent (water) can be calculated as:
\[ \text{Mass of solvent} = \text{Mass of solution} - \text{Mass of solute} \]
\[ \text{Mass of solvent} = 530 \text{ g} - 170 \text{ g} = 360 \text{ g} \]

### Step 7: Calculating Percentage W/V
To find the weight/volume percentage (W/V) of H₂O₂:
\[ \text{Percentage W/V} = \left( \frac{\text{Mass of H₂O₂}}{\text{Volume of solution (mL)}} \right) \times 100 \]
\[ \text{Percentage W/V} = \left( \frac{170 \text{ g}}{1000 \text{ mL}} \right) \times 100 = 17\% \]

### Step 8: Calculating Mole Fraction of H₂O₂
The mole fraction (χ) of H₂O₂ is calculated as follows:
- Moles of water (H₂O) = mass of water / molar mass of water = 360 g / 18 g/mol = 20 moles
- Total moles = moles of H₂O₂ + moles of H₂O = 5 + 20 = 25

Thus, the mole fraction of H₂O₂ is:
\[ \chi_{H₂O₂} = \frac{\text{Moles of H₂O₂}}{\text{Total moles}} = \frac{5}{25} = 0.2 \]

### Step 9: Calculating Molality
Molality (m) is defined as:
\[ m = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} \]
Converting the mass of solvent to kg:
\[ \text{Mass of solvent} = 360 \text{ g} = 0.36 \text{ kg} \]
Thus,
\[ m = \frac{5 \text{ moles}}{0.36 \text{ kg}} \approx 13.89 \text{ mol/kg} \]

### Conclusion
Based on the calculations:
- Molarity is approximately 5 mol/L (not 6, so option A is incorrect).
- Percentage W/V is 17% (option B is correct).
- Mole fraction is 0.2 (not 0.25, so option C is incorrect).
- Molality is approximately 13.89 mol/kg (matches option D).

### Final Answer
The correct options representing the concentration of the solution in other units are:
- Option B (17% W/V)
- Option D (molality as 1000/72)