The weight of a gaseous mixture containing `12.044xx10^(23)` atoms of He and `3.011xx10^(23)` molecules of hydrogen is _____________g.

Answer

To find the weight of a gaseous mixture containing \( 12.044 \times 10^{23} \) atoms of helium (He) and \( 3.011 \times 10^{23} \) molecules of hydrogen (H₂), we will follow these steps:

### Step 1: Calculate the number of moles of helium.
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{Number of atoms}}{N_A} \]
where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \) atoms/mole.

For helium:
\[ \text{Number of moles of He} = \frac{12.044 \times 10^{23}}{6.022 \times 10^{23}} = 2 \text{ moles} \]

### Step 2: Calculate the mass of helium.
The molar mass of helium is \( 4 \, \text{g/mol} \). Therefore, the mass of helium can be calculated as:
\[ \text{Mass of He} = \text{Number of moles} \times \text{Molar mass} = 2 \, \text{moles} \times 4 \, \text{g/mol} = 8 \, \text{g} \]

### Step 3: Calculate the number of moles of hydrogen.
For hydrogen (H₂), the number of moles can be calculated similarly:
\[ \text{Number of moles of H}_2 = \frac{\text{Number of molecules}}{N_A} \]
For hydrogen:
\[ \text{Number of moles of H}_2 = \frac{3.011 \times 10^{23}}{6.022 \times 10^{23}} = 0.5 \text{ moles} \]

### Step 4: Calculate the mass of hydrogen.
The molar mass of hydrogen (H₂) is \( 2 \, \text{g/mol} \). Therefore, the mass of hydrogen can be calculated as:
\[ \text{Mass of H}_2 = \text{Number of moles} \times \text{Molar mass} = 0.5 \, \text{moles} \times 2 \, \text{g/mol} = 1 \, \text{g} \]

### Step 5: Calculate the total mass of the gaseous mixture.
Now, we can find the total mass of the mixture by adding the masses of helium and hydrogen:
\[ \text{Total mass} = \text{Mass of He} + \text{Mass of H}_2 = 8 \, \text{g} + 1 \, \text{g} = 9 \, \text{g} \]

### Final Answer:
The weight of the gaseous mixture is \( 9 \, \text{g} \).

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