To find the mass of [tex]\( 5.30 \times 10^{22} \)[/tex] formula units of barium nitrate [tex]\(\left( Ba \left( NO _3\right)_2\right)\)[/tex], we need to follow these steps:
1. Determine the Number of Moles:
The first step is to convert the given number of formula units into moles. For this, we need Avogadro's number, which tells us how many entities (atoms, molecules, etc.) are in one mole. Avogadro's number is approximately [tex]\( 6.022 \times 10^{23} \)[/tex] entities per mole.
[tex]\[
\text{Number of moles} = \frac{\text{Number of formula units}}{\text{Avogadro's number}}
\][/tex]
Given:
[tex]\[
\text{Number of formula units} = 5.30 \times 10^{22}
\][/tex]
[tex]\[
\text{Avogadro's number} = 6.022 \times 10^{23}
\][/tex]
Plugging in the values:
[tex]\[
\text{Number of moles} = \frac{5.30 \times 10^{22}}{6.022 \times 10^{23}} \approx 0.0880 \text{ moles}
\][/tex]
2. Calculate the Mass:
Now that we have the number of moles, we can determine the mass of the barium nitrate by using the molar mass. The molar mass of [tex]\(Ba \left( NO _3\right)_2\)[/tex] is given as [tex]\(261.35 \, \text{g/mol}\)[/tex].
[tex]\[
\text{Mass} = \text{Number of moles} \times \text{Molar mass}
\][/tex]
Using the number of moles we calculated:
[tex]\[
\text{Mass} = 0.0880 \, \text{moles} \times 261.35 \, \text{g/mol} \approx 23.0 \, \text{g}
\][/tex]
Therefore, the mass of [tex]\( 5.30 \times 10^{22} \)[/tex] formula units of [tex]\( Ba \left( NO _3\right)_2 \)[/tex] is approximately [tex]\( 23.0 \, \text{g} \)[/tex].
So, the correct answer is:
[tex]\[ \boxed{23.0 \, \text{g}} \][/tex]
