Question:easy

The mass of one Avogadro number of helium atoms is:

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An Avogadro number of atoms is 1 mole; mass of 1 mole equals the atomic mass in grams.
Updated On: Jul 16, 2026
  • 1.00 g
  • 4.00 g
  • 8.00 g
  • 4 x 6.02 x 1023 g
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The Correct Option is B

Solution and Explanation

Start from what one Avogadro number of atoms actually means: it means exactly $N_A = 6.022 \times 10^{23}$ atoms, and by definition this exact count of atoms of any element is called one mole of that element.

  • The molar mass of an element, in grams per mole, is set equal to its atomic mass expressed in grams. This is the whole point of defining the mole the way chemists do.
  • Helium's atomic mass is 4, from 2 protons plus 2 neutrons in its common isotope He-4, so its molar mass is 4 g/mol.
  • Therefore, 1 mole of helium, that is $6.022 \times 10^{23}$ helium atoms, weighs 4.00 grams. No extra calculation is needed beyond reading off the atomic mass.
  • Option A, 1.00 g, would be true only for 1 mole of hydrogen atoms, not helium. Option D just rewrites the atom count multiplied by a leftover factor without simplifying it into a proper final mass.

So the correct mass of an Avogadro number of He atoms is 4.00 g, option B.

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