Which has maximum number of atoms:

Options

A. 24 of C (12)
B. 56 g of Fe (56)
C. 27 g of Al (27)
D. 108 g of Ag (108)

Explanation

To determine which option has the maximum number of atoms, we will calculate the number of atoms for each given mass of the elements. We will use the formula:

\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{atomic mass (g/mol)}} \]

And then use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mole) to find the total number of atoms.

### Step-by-Step Solution:

1. **Calculate the number of moles of Carbon (C)**:
- Given mass = 24 g
- Atomic mass of Carbon (C) = 12 g/mol
- Number of moles of C = \(\frac{24 \, \text{g}}{12 \, \text{g/mol}} = 2 \, \text{moles}\)
- Number of atoms in C = \(2 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 12.044 \times 10^{24} \, \text{atoms}\)

2. **Calculate the number of moles of Iron (Fe)**:
- Given mass = 26 g
- Atomic mass of Iron (Fe) = 56 g/mol
- Number of moles of Fe = \(\frac{26 \, \text{g}}{56 \, \text{g/mol}} = 0.4643 \, \text{moles}\)
- Number of atoms in Fe = \(0.4643 \, \text{moles} \times 6.022 \times 10^{23} \, \text{atoms/mole} \approx 2.79 \times 10^{23} \, \text{atoms}\)

3. **Calculate the number of moles of Aluminum (Al)**:
- Given mass = 27 g
- Atomic mass of Aluminum (Al) = 27 g/mol
- Number of moles of Al = \(\frac{27 \, \text{g}}{27 \, \text{g/mol}} = 1 \, \text{mole}\)
- Number of atoms in Al = \(1 \, \text{mole} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 6.022 \times 10^{23} \, \text{atoms}\)

4. **Calculate the number of moles of Silver (Ag)**:
- Given mass = 108 g
- Atomic mass of Silver (Ag) = 108 g/mol
- Number of moles of Ag = \(\frac{108 \, \text{g}}{108 \, \text{g/mol}} = 1 \, \text{mole}\)
- Number of atoms in Ag = \(1 \, \text{mole} \times 6.022 \times 10^{23} \, \text{atoms/mole} = 6.022 \times 10^{23} \, \text{atoms}\)

### Summary of Results:
- Carbon (C): \(12.044 \times 10^{24} \, \text{atoms}\)
- Iron (Fe): \(2.79 \times 10^{23} \, \text{atoms}\)
- Aluminum (Al): \(6.022 \times 10^{23} \, \text{atoms}\)
- Silver (Ag): \(6.022 \times 10^{23} \, \text{atoms}\)

### Conclusion:
The maximum number of atoms is found in **24 grams of Carbon (C)**, which has \(12.044 \times 10^{24} \, \text{atoms}\).