Respuesta :
The density of silver can be calculated by using the following formula:
[tex]d=\frac{m}{V}[/tex]
Here, m is mass and V is volume.
To calculate density, mass of silver needs to be calculated first.
In FCC, the number of Ag atoms in a unit cell will be 4, atomic mass of Ag is 107.87 g/mol and number of atoms in 1 mol are [tex]6.023\times 10^{23}[/tex] atoms, thus, mass can be calculated as follows:
[tex]m=4 atoms\times \frac{1 mol}{6.023\times 10^{23}atoms}\times \frac{107.87 g}{mol}=7.163\times 10^{-22} g[/tex].
Now, volume can be calculated as follows:
[tex]V=a^{3}[/tex]
First convert pm to cm:
[tex]1 pm=10^{-10} cm[/tex]
Thus,
[tex]408.7 pm=4.087\times 10^{-8} cm[/tex]
Putting the value to calculate volume,
[tex]V=(4.087\times 10^{-8} cm)^{3}=6.83\times 10^{-23} cm^{3}[/tex]
Now, calculate density as follows:
[tex]d=\frac{7.163\times 10^{-22}g}{6.627\times 10^{-23}cm^{3}}=10.5 g/cm^{3}[/tex]
Therefore, density of silver is [tex]10.5 g/cm^{3}[/tex].
the density of silver is calculated as
[tex]\rho = 10.49539 g/cm^3[/tex]
Data;
- edge length = 408.7pm
- atomic mass = 107.8682
- Avogadro's number = 6.022*10^23 atoms /mol
Density of Silver
The density of a substance is the ratio between the mass to the volume of that substance. In this case, the density of silver is calculated as
[tex]\rho = \frac{Zm}{a^3 N_a}[/tex]
Let's substitute the values into the formula
[tex]\rho = \frac{Zm}{a^3 N_a} \\\rho = \frac{4*107.862}{(4.087*10^-^8)^3 * 6.023*10^2^2}\\ \rho = 10.49539 g/cm^3[/tex]
From the calculations above, the density of silver is calculated as
[tex]\rho = 10.49539 g/cm^3[/tex]
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