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 $0.450~\mathrm{M}$.


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.


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