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

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

Step-by-step 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

Answer 2

The number of Aluminium atoms present in per cubic meter is
`N=6.02* 10^(28)\ atoms/m^3`

Step-by-step 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* 10^(23)\ atoms/mol`

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

`N=(\rho N_(A))/(A_(Al))`

Plug the given values in the formula we get,

`N=(2.70* 6.022* 10^(23))/(26.98)\\ \\N=0.602* 10^(23)\ atoms/cm^3\\N=6.02* 10^(22)\ atoms/cm^3\\N=6.02* 10^(22)*10^6\ atoms/m^3`

`N=6.02* 10^(28)\ atoms/m^3`

So, the number of Aluminium atoms present in per cubic meter is
`N=6.02* 10^(28)`