The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following element contains the greatest number of atom?

Options

A. 4g He
B. 46g Na
C. 0.40g Ca
D. 12 g He

Explanation

To determine which element contains the greatest number of atoms when given in moles, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Avogadro's Number**:
- Avogadro's number (Na) is \(6.022 \times 10^{23}\), which represents the number of atoms in one mole of any substance.

2. **Identify the Elements and Their Given Weights**:
- We need to compare the number of atoms in the following elements:
- Helium (He) - 4 g
- Sodium (Na) - 46 g
- Calcium (Ca) - 0.40 g
- Helium (He) - 12 g

3. **Calculate the Number of Moles for Each Element**:
- The formula for calculating the number of moles is:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} \]

- **For Helium (4 g)**:
- Given weight = 4 g
- Molecular weight of He = 4 g/mol
- Number of moles = \( \frac{4 \text{ g}}{4 \text{ g/mol}} = 1 \text{ mole} \)

- **For Sodium (46 g)**:
- Given weight = 46 g
- Molecular weight of Na = 23 g/mol
- Number of moles = \( \frac{46 \text{ g}}{23 \text{ g/mol}} = 2 \text{ moles} \)

- **For Calcium (0.40 g)**:
- Given weight = 0.40 g
- Molecular weight of Ca = 40 g/mol
- Number of moles = \( \frac{0.40 \text{ g}}{40 \text{ g/mol}} = 0.01 \text{ moles} \)

- **For Helium (12 g)**:
- Given weight = 12 g
- Molecular weight of He = 4 g/mol
- Number of moles = \( \frac{12 \text{ g}}{4 \text{ g/mol}} = 3 \text{ moles} \)

4. **Calculate the Number of Atoms for Each Element**:
- The number of atoms can be calculated by multiplying the number of moles by Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times N_a \]

- **For Helium (4 g)**:
- Number of atoms = \(1 \text{ mole} \times 6.022 \times 10^{23} = 6.022 \times 10^{23} \text{ atoms}\)

- **For Sodium (46 g)**:
- Number of atoms = \(2 \text{ moles} \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} \text{ atoms}\)

- **For Calcium (0.40 g)**:
- Number of atoms = \(0.01 \text{ moles} \times 6.022 \times 10^{23} = 6.022 \times 10^{21} \text{ atoms}\)

- **For Helium (12 g)**:
- Number of atoms = \(3 \text{ moles} \times 6.022 \times 10^{23} = 1.8066 \times 10^{24} \text{ atoms}\)

5. **Compare the Number of Atoms**:
- Helium (4 g): \(6.022 \times 10^{23}\) atoms
- Sodium (46 g): \(1.2044 \times 10^{24}\) atoms
- Calcium (0.40 g): \(6.022 \times 10^{21}\) atoms
- Helium (12 g): \(1.8066 \times 10^{24}\) atoms

6. **Conclusion**:
- The element that contains the greatest number of atoms is **12 g of Helium**, which has \(1.8066 \times 10^{24}\) atoms.