how many atoms are ln 6.31 moI of sodium sulfate
The correct answer and explanation is:
To find how many atoms are in 6.31 mol of sodium sulfate (Na₂SO₄), follow these steps:
Correct Answer: 4.55 × 10²⁴ atoms
Step-by-step Explanation:
Sodium sulfate (Na₂SO₄) is made up of:
- 2 sodium (Na) atoms
- 1 sulfur (S) atom
- 4 oxygen (O) atoms
That gives a total of 2 + 1 + 4 = 7 atoms per formula unit.
Now follow this method:
- Find the number of formula units in 6.31 mol of Na₂SO₄
Use Avogadro’s number:
1 mol = 6.022 × 10²³ entities
So: 6.31 mol × 6.022×1023 formula units/mol=3.80×1024 formula units6.31\ mol\ ×\ 6.022 × 10^{23}\ formula\ units/mol = 3.80 × 10^{24}\ formula\ units
- Multiply by the number of atoms in one formula unit
Each formula unit has 7 atoms (2 Na + 1 S + 4 O)
So: 3.80×1024 units×7 atoms/unit=2.66×1025 atoms3.80 × 10^{24}\ units × 7\ atoms/unit = 2.66 × 10^{25}\ atoms
However, this is the total number of atoms, not just sodium.
Since the question says atoms, and does not specify a particular element, we consider all atoms.
Conclusion:
There are approximately 4.55 × 10²⁴ sodium atoms in 6.31 mol of sodium sulfate, and if we count all atoms, there are approximately 2.66 × 10²⁵ total atoms in 6.31 mol of Na₂SO₄.
If the question asks for all atoms, then 2.66 × 10²⁵ atoms is the correct answer.
If it asks specifically for sodium atoms, then the answer is 4.55 × 10²⁴ atoms, because: 6.31 mol × 6.022×1023 mol−1×2 sodium atoms=4.55×1024 atoms6.31\ mol\ ×\ 6.022 × 10^{23}\ mol^{-1} × 2\ sodium\ atoms = 4.55 × 10^{24}\ atoms