Here's an alternative approach to keep in mind when dealing with dilutions.
As you know, the underlying principle of a dilution is that you can decrease the concentration of a solution by increasing its volume while keeping the number of moles of solute constant.
This implies that increasing the volume of the solution by a factor, let's say
This factor is called dilution factor and can be calculated like this
#color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))#
In your case, you know that the initial volume of the solution, i.e. the volume of the concentrated solution, is equal to
#0.500 color(red)(cancel(color(black)("L"))) * (10^3"mL")/(1color(red)(cancel(color(black)("L")))) = "500 mL"#
This means that the concentrated solution was diluted by a factor of
#"DF" = (500 color(red)(cancel(color(black)("mL"))))/(100color(red)(cancel(color(black)("mL")))) = color(blue)(5)#
As a result, the concentration of the diluted solution will be
#c_"diluted" = 1/color(blue)(5) * "0.0234 M"#
I make it 0.00468 M.
I tend to think of it like this:
The number of moles in the final dilution will be the same, but the total volume will change from 100 ml to 0.5 l .
So the molarity at the end will be 0.00234 / 0.5 which is 0.00468 M.