Answer:
The mass of ammonium nitrate that has the same number of nitrogen atoms is 5077.5 g.
Explanation:
To calculate the mass of ammonium nitrate we need to find the number of atoms in liquid nitrogen, as follows:
[tex]d_{N} = \frac{m_{N}}{V}[/tex]
Where:
[tex]d_{N}[/tex]: is the density of liquid nitrogen = 0.808 g/mL
[tex]m_{N}[/tex]: is the mass of liquid nitrogen
V: is the volume of liquid nitrogen = 2.2 L
By calculating the mass of liquid nitrogen we can find the number of atoms:
[tex]m_{N}=d_{N}*V = 0.808 g/mL*\frac{1000 mL}{1 L}*2.2 L=1777.6 g[/tex]
[tex]n = N_{A}*\eta_{N}[/tex]
[tex]n=N_{A}*\frac{m_{N}}{M_{N}}[/tex]
Where:
[tex]N_{A}[/tex]: is the Avogadro's number = 6.022x10²³ atoms/mol
n: is the number of atoms of liquid nitrogen
[tex]\eta_{N}[/tex]: is the number of moles of liquid nitrogen
[tex]M_{N}[/tex]: is the molar mass of liquid nitrogen = 28.014 g/mol
[tex]n = \frac{6.022 \cdot 10^{23} atoms}{1 mol}*\frac{1777.6 g}{28.014 g/mol} = 3.82 \cdot 10^{25} atoms[/tex]
Finally, the mass of ammonium nitrate is:
[tex]m_{NH_{4}NO_{3}} = \frac{n*M_{NH_{4}NO_{3}}}{N_{A}}[/tex]
Where:
[tex]M_{NH_{4}NO_{3}}[/tex]: is the molar mass of ammonium nitrate = 80.043 g/mol
[tex] m_{NH_{4}NO_{3}} = \frac{3.82 \cdot 10^{25} atoms*80.043 g/mol}{6.022 \cdot 10^{23} atoms/mol} = 5077.5 g [/tex]
Hence, the mass of ammonium nitrate that has the same number of nitrogen atoms is 5077.5 g.
I hope it helps you!
