To determine which element has 6.02 x 10^23 atoms in 4 grams, we need to calculate the number of moles of each element in 4 grams and then use Avogadro's number (6.02 x 10^23) to determine the number of atoms.
The molar mass of each element can be calculated using the atomic masses from the periodic table:
- The molar mass of helium (He) is 4.0026 g/mol
- The molar mass of lithium (Li) is 6.941 g/mol
- The molar mass of beryllium (Be) is 9.0122 g/mol
Using the formula for moles (moles = mass/molar mass), we can calculate the number of moles of each element in 4 grams:
- For helium: moles = 4 g / 4.0026 g/mol = 0.9996 mol
- For lithium: moles = 4 g / 6.941 g/mol = 0.5766 mol
- For beryllium: moles = 4 g / 9.0122 g/mol = 0.4435 mol
Now, we can use Avogadro's number to determine the number of atoms:
- For helium: 0.9996 mol x 6.02 x 10^23 atoms/mol = 6.02 x 10^23 atoms
- For lithium: 0.5766 mol x 6.02 x 10^23 atoms/mol = 3.47 x 10^23 atoms
- For beryllium: 0.4435 mol x 6.02 x 10^23 atoms/mol = 2.67 x 10^23 atoms
Therefore, the only element that has exactly 6.02 x 10^23 atoms in 4 grams is helium (4/2He). The answer is A.