# Chemistry

Hello,

I'm having some trouble getting started on this question:
"Calculate the volume and mass required to make 0.1 M solutions of Propionic Acid (1.0 M) and Sodium Propanoate (96.06 g/mol)"

I would normally not have trouble answering this question when either the mass or volume is provided but I do not know how begin answering the question

Any assistance would be greatly appreciated,

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1. The way I read the question is that there are two questions; i.e.,
1. How to prepare 0.1M solution of propionic acid and
2. How to prepare 0.1M solution of the salt.

1. Since no volume is given you just assume some volume you want to prepare. 1L is the easiest to assume although that could waste a lot of the compound. So you want 0.1 mols if you prepare 1L.
How much is 0.1mol? It's molar mass/10. Calculate that mass required. Then look up the density of proponic acid and use mass = volume x density to calculate the volume needed to provide that mass. Add to a volumetric flask, dissolve in a little H2O, make to the mark with DI water, shake thoroughly and stopper.

2. For sodium propionate, you want 0.1 mol to prepare 1L of the solution. 0.1 mol = mols mass in grams/10. Weigh that amount of the salt, place in a 1 L volumetric flask, dissolve in a little water, make to the mark with DI water, mix thoroughly, stopper.
I think the secret to answering this question is to decide just how much of each you wish to prepare.

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2. Thank you for your help
Just to be sure these were the answers I calculated:

Propionic Acid
Mass = 0.01g
Volume = 0.010ml

Sodium propionate
Mass = 0.9606g
Volume = I assume the is the same as the mass (please correct me if im wrong)

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3. I don't buy any of this. The numbers, even if they are right, (and they may be), are not realistic. Who would want to make 0.01 mL of such a solution? That's about two drops. :-)

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4. Yeah it did not seem to make any sense to me either.. I don't know what else I could possibly do in this problem

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5. I told you how to do it. You could try following my instructions. Here is how you do the sodium propionate.
You want 1 L of 0.1M CH3CH2COONa
mols needed = M x L = 0.1 M x 1 L = 0.1 mol
Then mol = grams/molar mass
0.1 mol = grams/96
grams = 96 x 0.1 = 9.6 grams.
Place 9.6 g the substance in a volumetric flask, dissolve in a little water, add DI water to the mark of the flask, mix thoroughly, stopper.

For the liquid propionic acid, You do the same for mols, then find the grams the same way, then you deviate from the above. Look up the density, calculate mL propionic acid (it's a liquid) from the density and the grams needed, place that many mL of the propionic acid in the volumetric flask and proceed as you did for the solid.

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