Question

How many atoms of aluminum are there in a piece of foil that has a volume of $2\text{ c}{{\text{m}}^{3}}$? The density of aluminum is $2.702g\text{ c}{{\text{m}}^{-3}}$.

Answer 1

First of all, we will find the mass of the aluminum atom by using the density formula as: $density=\dfrac{mass}{volume}$ and then, by using that mass we will find the moles using the unitary method and after finding the moles, apply the Avogadro’s law and you will get the number of atoms of the aluminum in the piece of foil. Now solve it.

Complete answer:
First of all, we will find the mass of the aluminum atom by using the formula as;
$Density=\dfrac{mass}{volume}$ ------------(1)
We know the density of the aluminum atom is= $2.702g\text{ c}{{\text{m}}^{-3}}$ (given)
volume of the aluminic atom=$2\text{ c}{{\text{m}}^{3}}$ (given)
Now put these values in equation (1), we get;
$\begin{aligned} & 2.702=\dfrac{mass}{2} \\\ & mass=2.702\times 2=5.404g \\\ \end{aligned}$
Thus, the mass of the aluminum atom is $5.404g$.
Now, we will convert the mass in gram into the moles as;
$26.982g$ of the aluminum atom contains=$1\text{ mole}$ of Al
Then,
$1g$of the aluminum atom will contain=$\dfrac{1}{26.982}\text{ moles}$of Al
And
$5.404g$of the aluminum atom will contain=$\dfrac{1}{26.982}\times \text{5}\text{.404 =0}\text{.2003 moles}$ of Al
Now, applying the Avogadro’s law:
$1\text{ mole}$ of the aluminum contains the= $6.023\times {{10}^{23}}$ atoms of Al
Then,
$\text{0}\text{.2003 moles}$ of the aluminum will contain= $\dfrac{6.023\times {{10}^{23}}}{0.2003}=1.21\times {{10}^{23}}$ atoms of Al
Thus, $1.21\times {{10}^{23}}$ atoms of aluminum are there in a piece of foil that has a volume of $2\text{ c}{{\text{m}}^{3}}$ and density is $2.702g\text{ c}{{\text{m}}^{-3}}$.

Note: Avogadro’s law states that the one mole of the gas at the standard conditions of the temperature and pressure contains Avogadro number of particles i.e. $6.023\times {{10}^{23}}$and occupies the volume of $22.4$ liters.