Relation between moles, molar mass and mass

You are asked to prepare a 0.8500 M solution of aluminum nitrate. You find that you have only 50.00 g of the solid.

What is the maximum volume of solution that you can prepare?

I'm confusing my molar masses with grams. The molar mass of Aluminum nitrate $$\ce{Al(NO3)3}$$ is 213.0 g. If I have 50.00 g of the entire solid $$\ce{Al(NO3)3}$$ does that mean I have 50.00 times the amount of 213.0 g or will I end up Multiplying 50.00 g of Aluminum nitrate by 0.8500 M, then dividing that by the MM of Aluminum Nitrate. But then I don't know how to set up the rest of my dimensional analysis to reach Liters.$$%edit$$

If you have 50 g of a compound that has a molar mass of 213 $\frac{\rm{g}}{\rm mol}$, then you have $\frac{50~\rm{g}}{213 \frac{\rm{ g}}{\rm mol}}=\frac{50}{213} \rm{mol}=0.234~\rm{mol}$ of the compound.

If you want to prepare a 0.8500 molar solution, that means the concentration you want is $0.85~\rm{M}=0.85\frac{\rm{mol}}{\rm{L}}$.

With those two numbers, the final answer should be apparent. (Just divide them and make sure the units work out!)

• Many thanks! I actually did my dimensional analysis this way before but I was unsure of my answer. It is a confidence booster to see other's punch out the same work and answer for the same problem. :) Commented Jun 22, 2015 at 17:45

You have $x = (50~\mathrm{g} / 213~\mathrm{g})$ moles of aluminum nitrate. You need some volume of a solution where the molarity of aluminum nitrate is $0.85$. If you had $181.05~\mathrm{g}$ of aluminum nitrate in 1 liter of solution, you'd be fine.

But you have $x$ moles. Thus, you need to scale the volume of solution accordingly.

Set $0.85~\mathrm{mol / L} =\frac{ x~\mathrm{mol}}{y~\mathrm{L}}$, using the value you calculate for $x$ as above, and solve for $y$, the solution volume in liters.

• Hi! We don't strictly answer homework questions. Instead, we tend to focus on the procedure of the solution. We give hints, not the direct answer, as it's not useful to anyone. Please edit your post in according to our policy. Commented Jun 22, 2015 at 18:09
• Can you illustrate how my answer is substantially different from the one immediately above it, from Curt F.? That will help me in the future. I appreciate it. Commented Jun 22, 2015 at 18:13
• I agree with Todd that our two answers have substantially similar content (+1!). I added the HW tag to the question, but since the questioner had actually invested some amount of thought and showed some existing work, I thought an answer similar to mine was merited. I think the same is true of Todd's answer. Commented Jun 22, 2015 at 18:18
• Your answer has less explanations and it doesn't cover anything more than Curt already does. There's not much wrong with it, though I recommend a bit of elaboration. Thank you. Commented Jun 22, 2015 at 18:18
• Todd's answer and mine were submitted within 1 minute of each other. Commented Jun 22, 2015 at 18:19