What is the total mass, in grams, of 3.01 × 10^{23} atoms of helium gas (He)?

• A. 2.00 g
• B. 0.50 g
• C. 1.00 g
• D. 4.00 g

Answer: (A)

Verified step by step guidance

1. Identify the given quantity: 3.01 \times 10^{23} atoms of helium (He).
2. Recall that the molar mass of helium (He) is approximately 4.00 grams per mole, and one mole contains Avogadro's number of atoms, which is 6.02 \times 10^{23} atoms.
3. Set up a proportion to find the mass corresponding to 3.01 \times 10^{23} atoms using the formula: \(\text{mass} = \left( \frac{\text{number of atoms}}{\text{Avogadro's number}} \right) \times \text{molar mass}\).
4. Substitute the values into the formula: \(\text{mass} = \left( \frac{3.01 \times 10^{23}}{6.02 \times 10^{23}} \right) \times 4.00\) grams.
5. Simplify the fraction and multiply by the molar mass to find the total mass of helium atoms in grams.