Question

The mass of one Avogadro number of helium atoms is:

• A. 1.00 g
• B. 4.00 g
• C. 8.00 g
• D. 4 x 6.02 x 10²³ g

Hint:

An Avogadro number of atoms is 1 mole; mass of 1 mole equals the atomic mass in grams.

Answer 1

The Correct Option is B

Step 1: One Avogadro number of anything means \(6.022 \times 10^{23}\) particles of that thing. When we take \(6.022 \times 10^{23}\) atoms of an element, we have exactly 1 mole of that element, by the very definition of the mole.

Step 2: The mass of 1 mole of an element, in grams, is numerically equal to its atomic mass. This is how atomic mass units are scaled to grams using the Avogadro constant.

Step 3: Helium has an atomic mass of 4 u, its nucleus has 2 protons and 2 neutrons, giving a mass number of 4.

Step 4: So 1 mole of helium atoms, which is \(6.022 \times 10^{23}\) He atoms, has a mass of 4 grams. It is not 1 g, that would be for hydrogen atoms, not 8 g, that would be double this amount, and it should not be left as \(4 \times 6.022\times10^{23}\) g because that just restates the atom count without converting it into an actual gram value.

Answer: One Avogadro number of He atoms has a mass of 4.00 g, option B.