Question

Calculate the number of atoms per cubic meter in aluminum. The density and atomic weight of aluminum are 2.70 g/cm3 and 26.98 g/mol respectively. The value of Avogadro's number is 6.022 x 1023 atoms/mol.

Answer 1

Answer:

The number of Aluminium atoms present in per cubic meter is $N=6.02\times 10^{28}\ atoms/m^3$

Explanation:

Given the density of Aluminium $(\rho )$ is $2.70\ g/cm^3$

And the atomic weight of Aluminium $(A_{Al})$ is $26.98\ g/mol$

Also, the Avogadro's number $(N_A)$ is $6.023\times 10^{23}\ atoms/mol$

We need to find the number of atoms per cubic meter $(N)$

$N=\frac{\rho N_{A}}{A_{Al}}$

Plug the given values in the formula we get,

$N=\frac{2.70\times 6.022\times 10^{23}}{26.98}\\ \\N=0.602\times 10^{23}\ atoms/cm^3\\N=6.02\times 10^{22}\ atoms/cm^3\\N=6.02\times 10^{22}\times10^6\ atoms/m^3$

$N=6.02\times 10^{28}\ atoms/m^3$

So, the number of Aluminium atoms present in per cubic meter is $N=6.02\times 10^{28}$

Answer 2

Answer:

We have 6.026 *10^28 Al atoms per cubic meter

Explanation:

Step 1: Data given

The density and atomic weight of aluminum are 2.70 g/cm3 and 26.98 g/mol respectively.

The value of Avogadro's number is 6.022 * 10^23 atoms/mol

Step 2: Calculate the number of atoms / cubic meter

N = (Na * ρal) / Mal

⇒ with Na = the number of Avogadro = 6.022 * 10^23 atoms / mol

⇒ with ρal = the density of aluminium = 2.70 g/cm³

⇒with Mal = the atomic weight of aluminium = 26.98g/mol

N = (6.022*10^23 *2.70) / 26.98

N = 6.026 *10^22 atoms / cm³ = 6.026 *10^28 atoms / m³

We have 6.026 *10^28 Al atoms per cubic meter