There are `6.02 xx 10^(22)` molecules each of `N_(2),O_(2)` and `H_(2)` which are mixed together at `760 mm` and `273 K`. The mass of the mixture in grams is `:`

Options

A. `6.2`
B. `4.12`
C. `3.09`
D. 7

Explanation

To solve the problem step by step, we will calculate the mass of the mixture of gases (N₂, O₂, and H₂) given the number of molecules of each gas, the pressure, and the temperature.

### Step 1: Determine the number of moles of each gas
We are given that there are \(6.02 \times 10^{22}\) molecules of each gas. We know that Avogadro's number is approximately \(6.023 \times 10^{23}\) molecules per mole.

To find the number of moles (\(n\)) of each gas, we use the formula:

\[ n = \frac{\text{Number of molecules}}{\text{Avogadro's number}} \]

Substituting the values:

\[ n = \frac{6.02 \times 10^{22}}{6.023 \times 10^{23}} \approx 0.1 \text{ moles} \]

### Step 2: Calculate the molar mass of each gas
Next, we need to find the molar mass of each gas:

- Molar mass of \(N_2\) (Nitrogen) = 28 g/mol
- Molar mass of \(O_2\) (Oxygen) = 32 g/mol
- Molar mass of \(H_2\) (Hydrogen) = 2 g/mol

### Step 3: Calculate the mass of each gas
Now, we can calculate the mass of each gas using the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

For each gas:

1. Mass of \(N_2\):
\[ \text{Mass}_{N_2} = 0.1 \text{ moles} \times 28 \text{ g/mol} = 2.8 \text{ g} \]

2. Mass of \(O_2\):
\[ \text{Mass}_{O_2} = 0.1 \text{ moles} \times 32 \text{ g/mol} = 3.2 \text{ g} \]

3. Mass of \(H_2\):
\[ \text{Mass}_{H_2} = 0.1 \text{ moles} \times 2 \text{ g/mol} = 0.2 \text{ g} \]

### Step 4: Calculate the total mass of the mixture
Now, we can find the total mass of the mixture by adding the masses of all three gases:

\[ \text{Total mass} = \text{Mass}_{N_2} + \text{Mass}_{O_2} + \text{Mass}_{H_2} \]
\[ \text{Total mass} = 2.8 \text{ g} + 3.2 \text{ g} + 0.2 \text{ g} = 6.2 \text{ g} \]

### Final Answer
The mass of the mixture in grams is \(6.2 \text{ g}\).

---