How many oxygen atoms are present in one mole of sodium chromate (Na₂CrO₄)?

A. [tex]\( 2.4 \times 10^{24} \)[/tex] atoms
B. [tex]\( 4.2 \times 10^{24} \)[/tex] atoms
C. [tex]\( 1.2 \times 10^{24} \)[/tex] atoms
D. [tex]\( 6.0 \times 10^{23} \)[/tex] atoms



Answer :

To determine the number of oxygen atoms present in one mole of sodium chromate (Na₂CrO₄), we need to follow these steps:

1. Identify the number of oxygen atoms in one molecule of sodium chromate:
The molecular formula of sodium chromate (Na₂CrO₄) indicates that each molecule contains 4 oxygen atoms.

2. Use Avogadro's number:
Avogadro's number, [tex]\( 6.022 \times 10^{23} \)[/tex], represents the number of units (atoms, molecules, etc.) in one mole of a substance.

3. Calculate the number of oxygen atoms in one mole of sodium chromate:
Since one molecule of Na₂CrO₄ contains 4 oxygen atoms, one mole of Na₂CrO₄ will contain 4 times Avogadro's number of oxygen atoms.

Thus,
[tex]\[ \text{Number of oxygen atoms} = 4 \times 6.022 \times 10^{23} \][/tex]
Multiplying these together gives us:
[tex]\[ \text{Number of oxygen atoms} = 2.4088 \times 10^{24} \][/tex]

4. Compare with the given options:
The number of oxygen atoms in one mole of sodium chromate is [tex]\( 2.4088 \times 10^{24} \)[/tex], which closely matches one of the given options:

- [tex]\( 2.4 \times 10^{24} \)[/tex] atoms
- [tex]\( 4.2 \times 10^{24} \)[/tex] atoms
- [tex]\( 1.2 \times 10^{24} \)[/tex] atoms
- [tex]\( 6.0 \times 10^{23} \)[/tex] atoms

From the above calculation, we see that [tex]\( 2.4 \times 10^{24} \)[/tex] atoms is approximately correct.

Therefore, the closest answer is:
[tex]\[ 2.4 \times 10^{24} \text{ oxygen atoms} \][/tex]