# Question b7819

Jul 30, 2017

$1.8 \cdot {10}^{25}$

#### Explanation:

For starters, you know that aluminium has an atomic number of $13$, which means that a neutral atom of aluminium contains $13$ protons inside its nucleus and $13$ electrons surrounding the nucleus.

In order to become a $3 +$ cation, a neutral atom of aluminium must lose $3$ electrons. This means that every ${\text{Al}}^{3 +}$ cation contains $10$ electrons surrounding its nucleus.

Now, you can safely assume that the mass of an aluminium cation is equal to the mass of a neutral atom of aluminium.

This implies that you can use the molar mass of aluminium to calculate the number of moles of aluminium cations present in your sample

81 color(red)(cancel(color(black)("g"))) * "1 mole Al"^(3+)/(27.0color(red)(cancel(color(black)("g")))) = "3.0 moles Al"^(3+)

As you know, in order to be able to say that you have $1$ mole of aluminium cations, you need to have $6.022 \cdot {10}^{23}$ aluminium cations in your sample.

This means that your sample will contain

3.0 color(red)(cancel(color(black)("moles Al"^(3+)))) * overbrace((6.022 * 10^(23)color(white)(.)"Al"^(3+)color(white)(.)"cations")/(1color(red)(cancel(color(black)("mole Al"^(3+))))))^(color(blue)("Avogaro's constant")) = 1.81 * 10^(24)# ${\text{Al}}^{3 +}$ $\text{cations}$

Finally, use the fact that every aluminium cation contains $10$ electrons to say that your sample contains

$1.81 \cdot {10}^{24} \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{Al"^(3+)"cations"))) * "10 e"^(-)/(1color(red)(cancel(color(black)("Al"^(3+)"cations")))) = color(darkgreen)(ul(color(black)(1.8 * 10^(25)color(white)(.)"e}}^{-}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of ${\text{Al}}^{3 +}$.