Question

The number of atoms present in one mole of an element is equal to Avogadro number. Which of the following elements contains the greatest number of atoms?
a.) 4 g He
b.) 46 g Na
c.) 0.4 g Ca
d.) 12 g He

Answer 1

Hint: Avogadro number ($$N_A$$) representing the number of atoms, molecules, or ions present in one mole of the given substance. The number of atoms present in one mole of an element is equal to Avogadro number.
Avogadro number ($$N_A$$) = $$6.023\times {10}^{23}$$atoms
Complete answer:
We know that number atoms = moles × $$N_A$$ $$N_A$$= Avogadro number
Therefore number of moles $$={\displaystyle \frac{\text{weight in grams}}{\text{molecular weight}}}$$
We have to calculate the number atoms from the given options.
Coming to given options, Option A, 4 g of He
4 g of He $$={\displaystyle \frac{4}{4}}=1\text{ mole}$$
4 g of Helium is equal to 1 mole.
Option B, 46 g of Na $$={\displaystyle \frac{46}{23}}=2\text{ moles}$$
46 g of sodium is equal to 2 moles.
Option C, 0.4 g of Ca$$={\displaystyle \frac{0.40}{40}}=0.01\text{ mole}$$
0.4 g of Calcium is equal to 0.01 mole
Option D, 12 g of He $$={\displaystyle \frac{12}{4}}=3\text{ moles}$$
12 g of Helium is equal to 3 moles.
Hence, 12 g of He contains the greatest number of atoms.
So, the correct option is D.


Note: Don’t be confused with the words number of atoms and number of moles.
In one mole of given substance, Avogadro number ($$N_A$$) of atoms are present.
The relationship between number of moles and number of atoms is as follows.
One mole of a substance = $$6.023\times {10}^{23}$$atoms.