`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 find the total pressure exerted by the mixture of gases in the flask, we can 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: Calculate the number of moles of each gas.

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

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

3. **For Nitric Oxide (NO)**:
- Mass = 3 g
- Molar mass of NO = 14 (N) + 16 (O) = 30 g/mol
- Number of moles of NO = \( \frac{3 \text{ g}}{30 \text{ g/mol}} = 0.1 \text{ moles} \)

### Step 2: Calculate the total number of moles (n).

\[ n = n_{H2} + n_{He} + n_{NO} = 0.5 + 0.5 + 0.1 = 1.1 \text{ moles} \]

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

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

Substituting the values into the Ideal Gas Law:

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

Calculating the right side:

\[ P \cdot 1.1 = 1.1 \cdot 0.0821 \cdot 300 \]

\[ P \cdot 1.1 = 24.63 \text{ atm} \]

### Step 4: Solve for P.

\[ P = \frac{24.63 \text{ atm}}{1.1 \text{ L}} = 24.63 \text{ atm} \]

Thus, the total pressure exerted by the mixture is approximately **24.63 atm**.

### Final Answer:
The total pressure exerted by the mixture is **24.63 atm**.

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