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Use the periodic table to determine how many atoms of phosphorus (P) are in a sample that has a mass of [tex]172.90 \, \text{g}[/tex].

A. [tex]3.48 \times 10^{21}[/tex] atoms
B. [tex]1.97 \times 10^{22}[/tex] atoms
C. [tex]3.36 \times 10^{24}[/tex] atoms
D. [tex]1.04 \times 10^{26}[/tex] atoms

Sagot :

Sure, I'd be happy to help you figure this out!

To determine the number of atoms of phosphorus in a sample with a mass of [tex]\(172.90 \, \text{g}\)[/tex], we need to follow these steps:

1. Determine the molar mass of phosphorus:
The atomic mass of phosphorus is [tex]\(30.974 \, \text{g/mol}\)[/tex]. This value is obtained from the periodic table.

2. Calculate the number of moles of phosphorus:
To find the number of moles, we use the formula:
[tex]\[ \text{moles of phosphorus} = \frac{\text{mass of sample}}{\text{molar mass of phosphorus}} \][/tex]
Plugging in the values, we get:
[tex]\[ \text{moles of phosphorus} = \frac{172.90 \, \text{g}}{30.974 \, \text{g/mol}} \approx 5.582 \, \text{moles} \][/tex]

3. Calculate the number of atoms of phosphorus:
We know that one mole of any substance contains [tex]\(6.022 \times 10^{23}\)[/tex] atoms (Avogadro's number). So, to find the number of atoms, we multiply the number of moles by Avogadro's number:
[tex]\[ \text{atoms of phosphorus} = \text{moles of phosphorus} \times 6.022 \times 10^{23} \, \text{atoms/mol} \][/tex]
Substituting the number of moles we estimated:
[tex]\[ \text{atoms of phosphorus} = 5.582 \, \text{moles} \times 6.022 \times 10^{23} \approx 3.362 \times 10^{24} \, \text{atoms} \][/tex]

4. Simplify to match the given options:
By simplifying the scientific notation to match the closest option:
[tex]\[ \text{atoms of phosphorus} \approx 3.36 \times 10^{24} \, \text{atoms} \][/tex]

From the choices given:

[tex]\[ 3.48 \times 10^{21} \, \text{atoms}, \quad 1.97 \times 10^{22} \, \text{atoms}, \quad 3.36 \times 10^{24} \, \text{atoms}, \quad 1.04 \times 10^{26} \, \text{atoms} \][/tex]

The correct answer is:

[tex]\[ \boxed{3.36 \times 10^{24}} \][/tex]