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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 $0.450~\mathrm{M}$.

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    $\begingroup$ Homework questions are not discouraged. Homework questions without thoughts into the problem and/or an attempt to solve said problem, on the other hand, are not only discouraged but not allowed (posting such a question will result in your question getting flagged/removed). That said, this homework problem looks good. $\endgroup$ – Breaking Bioinformatics Jun 23 '15 at 14:44
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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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  • $\begingroup$ 1. I suggest breaking lines for formulas, nested fractions look bad inline. 2. You'll hardly need to measure the necessary volume to complete 50 mL. Rather you'll just complete the volume in a volumetric flask, this allows you to wash the glass you used to measure the 1st volume properly and also accounts for cases where volumes don't add up. 3. What's with the tl;dr? There are no short/long versions, just one. $\endgroup$ – Molx Jun 24 '15 at 2:13
  • $\begingroup$ 1. I edited my answer and hope that it is more readable now. 2. OK from laboratory POV you are right but i was actually thinking from mathematical POV 3. tl;dr means "too long, didn't read". I removed it. Actually I tried to read the whole question 2-3 times but lost track in the midway and then i assumed the question and answered the question. Thanks for pointing out these things. $\endgroup$ – Osman Mamun Jun 24 '15 at 3:11

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