32 g of oxygen and 3.0 g of hydrogen are mixed and kept in a vessel at 760 mm pressure and `0^(@)C`. The total volume occupied by the mixture will be nearly

Options

A. 22.4 litres
B. 33.6 litre
C. 56.0 litres
D. 44.8 litres

Explanation

To solve the problem of finding the total volume occupied by a mixture of 32 g of oxygen and 3.0 g of hydrogen at a pressure of 760 mmHg and a temperature of 0°C, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure (in atm)
- \( V \) = Volume (in liters)
- \( n \) = Number of moles of gas
- \( R \) = Ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = Temperature (in Kelvin)

### Step 1: Convert the pressure from mmHg to atm
Since 1 atm = 760 mmHg, we can convert the pressure:

\[ P = \frac{760 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1 \text{ atm} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature in Kelvin can be calculated as follows:

\[ T = 0°C + 273 = 273 \text{ K} \]

### Step 3: Calculate the number of moles of each gas
To find the total number of moles, we need to calculate the moles of oxygen and hydrogen separately.

- **For Oxygen (O₂)**:
- Mass of oxygen = 32 g
- Molar mass of O₂ = 32 g/mol
- Moles of O₂ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mol}\)

- **For Hydrogen (H₂)**:
- Mass of hydrogen = 3 g
- Molar mass of H₂ = 2 g/mol
- Moles of H₂ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{2 \text{ g/mol}} = 1.5 \text{ mol}\)

### Step 4: Calculate the total number of moles
Now, we can find the total number of moles (\( n_T \)):

\[ n_T = \text{moles of O₂} + \text{moles of H₂} = 1 \text{ mol} + 1.5 \text{ mol} = 2.5 \text{ mol} \]

### Step 5: Use the Ideal Gas Law to find the volume
Now we can rearrange the Ideal Gas Law to solve for volume \( V \):

\[ V = \frac{nRT}{P} \]

Substituting the values we have:

- \( n = 2.5 \text{ mol} \)
- \( R = 0.0821 \text{ L·atm/(K·mol)} \)
- \( T = 273 \text{ K} \)
- \( P = 1 \text{ atm} \)

Calculating the volume:

\[ V = \frac{(2.5 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K})}{1 \text{ atm}} \]

Calculating this gives:

\[ V = \frac{(2.5) \times (0.0821) \times (273)}{1} \]

\[ V \approx 56.2 \text{ L} \]

Thus, the total volume occupied by the mixture is approximately **56 liters**.