What volume (in mL) of liquid `H_2O_2` has same number of molecules as there are number of molecules in 22.4 ml `H_2O(g)` at 2 atm and 546 K. [Given Density of liquid `H_2O_2=6.8xx 10^(-3)` gm/ml]

Answer

To find the volume of liquid \( H_2O_2 \) that has the same number of molecules as there are in 22.4 mL of \( H_2O(g) \) at 2 atm and 546 K, we can follow these steps:

### Step 1: Calculate the number of moles of \( H_2O(g) \)

Using the ideal gas equation:

\[ PV = nRT \]

Where:
- \( P = 2 \, \text{atm} \)
- \( V = 22.4 \, \text{mL} = 22.4 \times 10^{-3} \, \text{L} \)
- \( R = 0.082 \, \text{L atm/(K mol)} \)
- \( T = 546 \, \text{K} \)

Substituting the values into the equation:

\[ 2 \times (22.4 \times 10^{-3}) = n \times 0.082 \times 546 \]

Calculating the left side:

\[ 0.0448 = n \times 0.082 \times 546 \]

Now, calculate \( n \):

\[ n = \frac{0.0448}{0.082 \times 546} \]

Calculating the denominator:

\[ 0.082 \times 546 \approx 44.772 \]

Now, substituting back:

\[ n \approx \frac{0.0448}{44.772} \approx 1 \times 10^{-3} \, \text{moles} \]

### Step 2: Calculate the mass of \( H_2O_2 \)

The molar mass of \( H_2O_2 \) is approximately 34 g/mol. Therefore, the mass of \( H_2O_2 \) corresponding to \( 1 \times 10^{-3} \) moles is:

\[ \text{Mass of } H_2O_2 = n \times \text{Molar Mass} = 1 \times 10^{-3} \times 34 = 0.033966 \, \text{grams} \]

### Step 3: Calculate the volume of \( H_2O_2 \)

Using the density of \( H_2O_2 \):

\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]

Rearranging gives:

\[ \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

Substituting the values:

\[ \text{Density of } H_2O_2 = 6.8 \times 10^{-3} \, \text{g/mL} \]

Now, substituting the mass and density:

\[ \text{Volume} = \frac{0.033966}{6.8 \times 10^{-3}} \approx 5 \, \text{mL} \]

### Final Answer:

The volume of liquid \( H_2O_2 \) that has the same number of molecules as there are in 22.4 mL of \( H_2O(g) \) at 2 atm and 546 K is approximately **5 mL**.

---