I have a couple chemistry questions for any willing participants. Im reviewing for an exam and just want to make sure I have the right answers so here it goes:

What is the molarity of potassium acetate a solution prepared from 25 g potassium acetate and 175 g water? The density of water is 1 g/ml. Density of potassium acetate is 1.57 g/ml. My answer was 1.336m.

What is molality of above question? my answer was 1.457m.

Citric acid is frequently added to cleaning agents as a rust remover. A 5 g sample is added to 100g of water and the solution freezes at -.484 C. The freezing constant of water is 1.858 C what is the molar mass of citric acid? my answer was 192 g/mol

Molarity is abbreviated with a capital ‘M’. Molality is ‘m’. Check your rounding on #2.

When figuring molar mass using freezing-point depression, remember that your mass fraction, w, is the ratio of the mass of your solute divided by the total mass of the solution.

[quote]tedro wrote:
Molarity is abbreviated with a capital ‘M’. Molality is ‘m’. Check your rounding on #2.

When figuring molar mass using freezing-point depression, remember that your mass fraction, w, is the ratio of the mass of your solute divided by the total mass of the solution.[/quote]

I am confused about #3. This is what I did:

.484/1.858 = .260m

.260m x .1 kg water = .026mol

5g citric acid / .026mol = 192 g/mol

what answer are you getting?

I also just redid number 2 and got a slightly different answer ( i am kind of brain dead right about now)

First I find molar mass of potassium acetate

25g/98mol = .255 g/mol

then I find volume of solutions

Potassium: v=25gx1.57 = 39.25 ml
H20: v=175x1 = 175ml
Volume of solution = 175+39.25 = 214.25
Molarity: .255mol/.21425 L = 1.19 M

You were basically right the first time. Molality is mol/kg. You figured the number of moles correctly, but the kg is for the solvent only, not the solution. Therefore you need to divide .255 by 175, the mass of the water. You did this the first time and basically got the correct answer, your only problem was in rounding the .255.

Never round until your final answer, it messes up your significant digits. You should get 1.458, but since you only have two significant digits in your 25g, you should round to 1.5. (Or 1.46 if this was actually listed as 25.0g)

Your formula relating molar mass to freezing point depression is

M=[wK(sub f)]/(Delta T)

M is molar mass.
w is the mass fraction.
K(sub f) is your freezing consant of water, 1.858 C. (It is technically a bit different once the solute is introduced, but that is negligible and we don’t need to worry about it now.
Delat T is the difference between the freezing point of pure H2O and the freezing point of the solution, so it is .484 C.

To find w you divide the mass of the solute by the mass of the solution, so you have 5g/105g=.0477.

After this you can just plug every thing else in to get 182 g/mol.

*I am a bit anal about signigicant digits, and I don’t know how your professor grades with them. Based on the info you gave, your first answer should be 1.3 M, second should be 1.5 m, and the third should just be 200 g/mol, since there is only one significant digit in the 5g you listed. Of course you may have just left off the 5.00g.

Again, this is up to your professor but has always been a pet peave of mine. Significant digits seem to get largely ignored, but rounding error becomes pretty significant after a while.

I justed wanted to add something you seem to be missing.

Solute is the substance being disolved.

Solvent is the substance in which a solute is disolved. Water is a universal solvent.

A solution is made of both a solvent and a solute.

Molarity is mol/L of solution.

Molality is mol/kg of solvent.

Mass fraction is the ratio of solvent mass to solution mass.

When you tried #2 the second time, not only did you mistakenly add in the volume ot the potassium acetate, you also miscalculated it. You should have divided to get 15.9. You obviously did this correctly is solving #1.

The second time i calculated it I got number 1 and 2 mixed up which is what caused me getting the wrong answer. It was just a matter of unorganization. But as I sit here looking at number 1 right now i am unsure how I managed to get the right answer, I cant seem to get the right answer for myself at the moment.

EDIT:

Nevermind I figured it out once again. Its the easy questions that are tricking me. Thanks alot for your help thus far. As far as sig figs go, my teacher just wants 3 sig figs for the answer.

Here is two more if you feel like making sure I did them right:

In an experiment, the initial concentration of SO2Cl2 was .0248 mol/L. If the rate constant is 2.2 x 10^-5 /s at 25 C, what is the concentration of SO2Cl2 after 4.5 hours? Reaction is first order. My answer: .0174 mol/L

This one I am not too sure about:
Calculate the rate constant at 102 C for the decomposition of dinitrogen pentoxide given that the rate constant is 2.5x10^-3 at 59 C. The activation energy was calculated to be 100 kj/mol and R= 8.314 j/mol K. My answer was .00251 and i used the formula:

[quote]playmaker08 wrote:
The second time i calculated it I got number 1 and 2 mixed up which is what caused me getting the wrong answer. It was just a matter of unorganization. But as I sit here looking at number 1 right now i am unsure how I managed to get the right answer, I cant seem to get the right answer for myself at the moment.[/quote]

First you need to find the volume of the potassium acetate.

25g/1.57g/ml=15.9 ml potassium acetate.

15.9 ml + 175 ml = 190.9 ml solution. (Not exactly, due to some advanced chemistry that we won’t get into, but this is fine for now.

Damnation Ted, you didn’t leave anything for me to answer! You mathematicians are really good at chemistry. I wasn’t aware that you took any chemistry besides the required credits. You must be really reaching for mental stimulation these days :).

M = moles of solute/litres of solution, am I right? Using dimensional analysis, find a way to get the M of your solute from g. Do the same for molality. I have no idea how to solve the third one.

[quote]Aragorn wrote:
Damnation Ted, you didn’t leave anything for me to answer! You mathematicians are really good at chemistry. I wasn’t aware that you took any chemistry besides the required credits. You must be really reaching for mental stimulation these days :).

[/quote]

You forget that I was pre-med most of my time in school!

I took chem 1 & 2, organic chem 1 & 2, and even gen. phys. chem.

It was when I took phys chem that I realized I was much better at math than chemistry. I never really had a clue what was going on, but as long as I was able to keep variables and definitions straight, it was all just a little bit of calculus.

I will admit, I had to look some things up to answer the OP’s questions. I remembered the molarity/molality stuff off of the top of my head, but the rest took a little bit of work. I just can’t resist the challenge to prove that I can still do it!