`1g H_(2), 2g` He and 3g NO are contained in 1.1 L flask at 300 K. Total pressure exerted by the mixture is :

Options

A. 5.45 atm
B. 6.0 atm
C. 24.63 atm
D. 134.34 atm

Explanation

To solve the problem, we will use the Ideal Gas Law, which is represented 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) or 0.08 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of each gas.**

1. **For Hydrogen (H₂)**:
- Given mass = 1 g
- Molar mass of H₂ = 2 g/mol
- Number of moles (n) = mass / molar mass = \( \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ mol} \)

2. **For Helium (He)**:
- Given mass = 2 g
- Molar mass of He = 4 g/mol
- Number of moles (n) = mass / molar mass = \( \frac{2 \text{ g}}{4 \text{ g/mol}} = 0.5 \text{ mol} \)

3. **For Nitric Oxide (NO)**:
- Given mass = 3 g
- Molar mass of NO = 30 g/mol
- Number of moles (n) = mass / molar mass = \( \frac{3 \text{ g}}{30 \text{ g/mol}} = 0.1 \text{ mol} \)

**Step 2: Calculate the total number of moles (n_total) of the gas mixture.**

\[ n_{\text{total}} = n_{H_2} + n_{He} + n_{NO} = 0.5 + 0.5 + 0.1 = 1.1 \text{ mol} \]

**Step 3: Use the Ideal Gas Law to find the total pressure (P).**

- Given:
- Volume (V) = 1.1 L
- Temperature (T) = 300 K
- Ideal gas constant (R) = 0.0821 L·atm/(K·mol)

Using the Ideal Gas Law:

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

Substituting the values:

\[ P = \frac{(1.1 \text{ mol}) \times (0.0821 \text{ L·atm/(K·mol)}) \times (300 \text{ K})}{1.1 \text{ L}} \]

Calculating the pressure:

\[ P = \frac{(1.1) \times (0.0821) \times (300)}{1.1} = 24.3 \text{ atm} \]

### Final Answer:
The total pressure exerted by the mixture is approximately **24 atm**.