Calculate the number of moles of `Cl^(-)` ions in 1.75 L of `1.0 xx 10^-3` M `AlCl_3`. How would you solve this?

For starters, calculate how many moles of aluminium chloride would be needed in order to get a `1.0 * 10^(-3)` `"M"` solution in `"1.75 L"` of solution.

As you know, molarity tells you the number of moles of solute present in `"1 L"` of solution. This means that `"1 L"` of `1.0 * 10^(-3)` `"M"` aluminium chloride solution will contain `1.0 * 10^(-3)` moles of aluminium chloride, the solute.

In your case, the solution will contain

`1.75 color(red)(cancel(color(black)("L solution"))) * overbrace((1.0 * 10^(-3)color(white)(.)"moles AlCl"_3)/(1color(red)(cancel(color(black)("L solution")))))^(color(blue)(=1.0 * 10^(-3)"M AlCl"_3)) = 1.75 * 10^(-3)` `"moles AlCl"_3`

So, you know that you must dissolve `1.75 * 10^(-3)` moles of aluminium chloride in enough water to make the volume of the solution equal to `"1.75 L"`.

Now, aluminium chloride is a soluble ionic compound, which implies that it dissociates completely in aqueous solution to produce aluminium actions and chloride anions

`"AlCl"_ (color(red)(3)(aq)) -> "Al"_ ((aq))^(3+) + color(red)(3)"Cl"^(-)`

Notice that every `1` mole of aluminium chloride dissolved in water produces `color(red)(3)` moles of chloride anions.

This implies that your solution will contain

`1.75 * 10^(-3) color(red)(cancel(color(black)("moles AlCl"_3))) * (color(red)(3)color(white)(.)"moles Cl"^(-))/(1color(red)(cancel(color(black)("mole AlCl"_3))))`

`= color(darkgreen)(ul(color(black)(5.3 * 10^(-3)color(white)(.)"moles Cl"^(-))))`

The answer must be rounded to two sig figs, the number of sig figs you have for the molarity of the solution.