Discussion
The mass of one Avogadro number of helium atom is
Answer: D
One Avogadro number of helium atoms weighs 4.00 g.
By definition, \(N_A=6.022\times10^{23}\) atoms is one mole, and one mole of an element weighs its gram atomic mass. Helium's atomic mass is 4, so \[m = 1\ \text{mol}\times 4\ \mathrm{g\,mol^{-1}} = 4.00\ \mathrm{g}\] Multiplying 4 by \(6.02\times10^{23}\) would give the mass of that many moles, not of that many atoms.
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