The isotopic abundance of `C-12` and `C-14 "is" 98%` and `2%` respectively. What would be the number of `C-14` isotope in `12 g` carbon sample?

Options

A. `1.032xx10^(22)`
B. `3.01xx10^(23)`
C. `5.88xx10^(23)`
D. `6.02xx10^(23)`

Explanation

To find the number of `C-14` isotopes in a `12 g` carbon sample with isotopic abundances of `C-12` being `98%` and `C-14` being `2%`, we can follow these steps:

### Step 1: Calculate the mass of `C-14` in the `12 g` sample.
Given that the abundance of `C-14` is `2%`, we can find the mass of `C-14` in `12 g` of carbon.

\[ \text{Mass of } C-14 = \left(\frac{2}{100}\right) \times 12 \, \text{g} = 0.24 \, \text{g} \]

### Step 2: Calculate the number of moles of `C-14`.
To find the number of moles of `C-14`, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of `C-14` is `14 g/mol`. Thus,

\[ \text{Number of moles of } C-14 = \frac{0.24 \, \text{g}}{14 \, \text{g/mol}} = 0.01714 \, \text{mol} \]

### Step 3: Calculate the number of atoms of `C-14`.
Using Avogadro's number, which is approximately \(6.022 \times 10^{23}\) atoms/mol, we can find the number of atoms of `C-14`.

\[ \text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's number} \]

\[ \text{Number of atoms of } C-14 = 0.01714 \, \text{mol} \times 6.022 \times 10^{23} \, \text{atoms/mol} \approx 1.032 \times 10^{22} \, \text{atoms} \]

### Final Answer:
The number of `C-14` isotopes in a `12 g` carbon sample is approximately \(1.032 \times 10^{22}\) atoms.

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