The number of atoms present in a 0.635 g of Cu piece will be :

Options

A. `6.023 xx 10^(-23)`
B. `6.023 xx 10^(23)`
C. `6.023 xx 10^(22)`
D. `6.023 xx 10^(21)`

Explanation

To find the number of atoms in a 0.635 g piece of copper (Cu), we can follow these steps:

### Step 1: Calculate the number of moles of copper.
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of copper (Cu) is approximately 63.5 g/mol.

\[ \text{Number of moles of Cu} = \frac{0.635 \, \text{g}}{63.5 \, \text{g/mol}} \approx 0.01 \, \text{moles} \]

### Step 2: Use Avogadro's number to find the number of atoms.
Avogadro's number is \(6.022 \times 10^{23}\) atoms/mol, which tells us how many atoms are in one mole of a substance.

To find the number of atoms in 0.01 moles of copper, we multiply the number of moles by Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times \text{Avogadro's number} \]
\[ \text{Number of atoms} = 0.01 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} \]
\[ \text{Number of atoms} \approx 6.022 \times 10^{21} \, \text{atoms} \]

### Final Answer:
The number of atoms present in a 0.635 g piece of copper is approximately \(6.022 \times 10^{21}\) atoms.

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