What is the concept behind the calculations for the dilution of solutions?

Question

I'm interested in the concept behind this computation and was wondering if anyone could give me some pointers or hints. I've prepared $0.28~\mathrm{L}$ solution of $0.8500~\mathrm{M}$ of $\ce{Al(NO3)3}$. I need a $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$ of $\ce{Al(NO3)3}$ made from my original solution.

For my dimensional analysis what I need should be up front right? $50~\mathrm{mL}$ of the solution first converted to liters for the bridge conversion to 0.450 moles.

Once I have the molarity of my $50~\mathrm{mL}$ set to $0.450~\mathrm{mol/L}$ I divide my molarity by the molarity of the original solution I am taking this out of in the first place which gives me 0.026 liters of solution

$0.026~\mathrm{L}$ of solution is then what I need out of my $0.28~\mathrm{L}$ $0.8500~\mathrm{M}$ solution in order to come up with my 50 milliliters of $0.450~\mathrm{M}$ of solution? (I feel that something is missing)

My computations (conceptually explained above in my best effort)

$50~\mathrm{mL} \cdot 1~\mathrm{L}/1000~\mathrm{mL} \cdot 0.450~\mathrm{M}/1~\mathrm{L} \cdot 1~\mathrm{L}/0.8500~\mathrm{M} = 0.026~\mathrm{L}$ of $0.8500~\mathrm{M}$ solution needed to prepare $50~\mathrm{mL}$ solution of $0.450~\mathrm{M}$.

Answer 1

I'm not sure the exact question but I suppose you want to find the volume of 1st solution needed to come up with the 2nd solution. I always feel more comfortable with moles for this types of problem. Because mole is conserved (in a way).
So, to make the 2nd solution you need

$$50~\mathrm{ml} \times \frac{1~\mathrm{l}}{1000~\mathrm{ml}} \times 0.45~\mathrm{\frac {mol}{l}}=0.0225~\mathrm{mol}$$

That means you need $0.0225~\mathrm{mol}$ of $\ce{Al(NO3)3}$ from the first solution. You know the molarity and mole number. So the volume is $$\frac {0.0225~\mathrm{mol}}{0.85~\mathrm{\frac{mol}{l}}}=0.0264~\mathrm{l}=26.47~\mathrm{ml}.$$ But you need $50~\mathrm{ml}$ of the 2nd solution, so you need to add solvent (Water or whatever you used). The volume of extra solvent required is $50-26.47=23.53~\mathrm{ml}$. Now you got the 2nd solution and to check that you have got everything right, you can find the molarity of your new solution.