The number of molecules of `CO_(2)` liberated by the complete combustion of `0.1g` atoms of graphite in air is

Options

A. `3.01 xx 10^(22)`
B. `6.02 xx 10^(23)`
C. `6.02 xx 10^(22)`
D. `3.01 xx 10^(23)`

Explanation

To solve the problem of determining the number of molecules of \( CO_2 \) liberated by the complete combustion of \( 0.1 \) gram atoms of graphite, we can follow these steps:

### Step 1: Understand the concept of gram atoms and moles
1 gram atom of an element is equivalent to 1 mole of that element. Therefore, \( 0.1 \) gram atoms of graphite is equal to \( 0.1 \) moles of graphite.

### Step 2: Write the combustion reaction
The complete combustion of carbon (graphite) can be represented by the following equation:
\[ C + O_2 \rightarrow CO_2 \]
This indicates that 1 mole of carbon reacts with 1 mole of oxygen to produce 1 mole of carbon dioxide.

### Step 3: Determine the amount of \( CO_2 \) produced
Since \( 0.1 \) moles of graphite (carbon) are being combusted, and according to the stoichiometry of the reaction, \( 0.1 \) moles of carbon will produce \( 0.1 \) moles of \( CO_2 \).

### Step 4: Convert moles of \( CO_2 \) to molecules
To find the number of molecules, we use Avogadro's number, which states that 1 mole of any substance contains \( 6.022 \times 10^{23} \) molecules. Therefore, the number of molecules of \( CO_2 \) produced is:
\[ \text{Number of molecules} = \text{moles of } CO_2 \times \text{Avogadro's number} \]
\[ \text{Number of molecules} = 0.1 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mole} \]
\[ \text{Number of molecules} = 6.022 \times 10^{22} \, \text{molecules} \]

### Final Answer
The number of molecules of \( CO_2 \) liberated by the complete combustion of \( 0.1 \) gram atoms of graphite in air is \( 6.022 \times 10^{22} \) molecules.

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