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.