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 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-by-Step Solution:

1. **Convert the mass of gases to moles**:
- For oxygen (O₂):
- Molar mass of O₂ = 32 g/mol
- Moles of O₂ = \(\frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mol}\)

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

2. **Calculate the total number of moles (n)**:
- Total moles \( n = \text{moles of O₂} + \text{moles of H₂} = 1 + 1.5 = 2.5 \text{ mol}\)

3. **Convert the pressure from mmHg to atm**:
- Pressure in atm = \(\frac{760 \text{ mmHg}}{760 \text{ mmHg/atm}} = 1 \text{ atm}\)

4. **Convert the temperature from Celsius to Kelvin**:
- Temperature in Kelvin = \(0°C + 273 = 273 \text{ K}\)

5. **Substitute the values into the Ideal Gas Law**:
- Using the ideal gas equation \( PV = nRT \):
- \( V = \frac{nRT}{P} \)
- Substitute \( n = 2.5 \text{ mol} \), \( R = 0.0821 \text{ L·atm/(K·mol)} \), \( T = 273 \text{ K} \), and \( P = 1 \text{ atm} \):
\[ V = \frac{(2.5 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K})}{1 \text{ atm}} \]

6. **Calculate the volume**:
\[ V = \frac{(2.5) \times (0.0821) \times (273)}{1} \approx 56.03 \text{ L} \]

### Final Answer:
The total volume occupied by the mixture will be nearly **56.03 liters**.

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