Question #b7819

1 Answer
Jul 30, 2017

Answer:

#1.8 * 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 #"Al"^(3+)# cation contains #10# electrons surrounding its nucleus.

http://spmchemistry.onlinetuition.com.my/2013/10/formation-of-positive-ion.html

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 * 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)# #"Al"^(3+)# #"cations"#

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

#1.81 * 10^(24)color(red)(cancel(color(black)("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 #"Al"^(3+)#.