# Big Review (2 posts in 1)

How many grams of propane are in a 35L container at 40 deg Celsius and 900 torr?
40 deg C = 40 + 273K = 313 K

900 torr/768 torr = 1.2 atm (when do I convert to atm? I did it here simply to ease calculations)

Propane: C3H8 --> M.W. = 44.1 g
PV=nRT where R= .0821

Sol:
n=1.2 atm * 35L/.0821*313 = 1.6344 mol

?grams of C3H8= 1.63mol*(44.1g/1 mol C3H8) = 71.8 (have problem with significan figures; for example what criteria here dictates how many sig fig in the final answer,in this case to three sig figs?)

2. A solution of 20% ethanol and 80% water, by mass, has a density of 0.966 g/mL. Find the molality and the molarity of the ethanol in this solution.

Molality = #mol solute/#kg solvent
Molarity = #mol/L (solution)

.966g/mL * 1000mL = 966 grams solution

I ask this question: What is 20% of 966.
Sol: ? = .2 * 966
= 193.2g ethanol and the rest is water or 966 - 193.2 = 772.8 g H2O (this info not needed for M though)

193.2 g *(1 molC2H6O/46.1g)=4.19 mol

Molarity = 4.19 mol/L

Molaltity then is: what is 80% of 966

Sol: ? = .8 * 966 = 772.8 g (already determined)

convert: 772.8 g (1kg/1000 g)=.773kg (sig figs)
Molality then is:
4.19 molC2H6O/.773kg = 5.42 C2H6O m (molal)

How many grams of propane are in a 35L container at 40 deg Celsius and 900 torr?
40 deg C = 40 + 273K = 313 K

900 torr/768 torr = 1.2 atm (when do I convert to atm? I did it here simply to ease calculations) This is the best way to do it. You are right, it is easy and avoids confusion.However, note that you had three significant figures in all the numbers so you should not have rounded the answer to 1.2. It should be 1.18 atm.

Propane: C3H8 --> M.W. = 44.1 g
PV=nRT where R= .0821

Sol:
n=1.2 atm * 35L/.0821*313 = 1.6344 mol

?grams of C3H8= 1.63mol*(44.1g/1 mol C3H8) = 71.8 (have problem with significan figures; for example what criteria here dictates how many sig fig in the final answer,in this case to three sig figs?) Except for the error above rounding 1.18 to 1.2, your calculation is ok. Keeping the 1.8 I get 71.0 grams to three significant figures. I am assuming the 35 L was 35.0 L; otherwise, two is right.

2. A solution of 20% ethanol and 80% water, by mass, has a density of 0.966 g/mL. Find the molality and the molarity of the ethanol in this solution.

Molality = #mol solute/#kg solvent
Molarity = #mol/L (solution)

.966g/mL * 1000mL = 966 grams solution

I ask this question: What is 20% of 966.
Sol: ? = .2 * 966
= 193.2g ethanol and the rest is water or 966 - 193.2 = 772.8 g H2O (this info not needed for M though)

193.2 g *(1 molC2H6O/46.1g)=4.19 mol

Molarity = 4.19 mol/L

Molaltity then is: what is 80% of 966

Sol: ? = .8 * 966 = 772.8 g (already determined)

convert: 772.8 g (1kg/1000 g)=.773kg (sig figs)
Molality then is:
4.19 molC2H6O/.773kg = 5.42 C2H6O m (molal)

This one is ok except that I would have written the zero in front of the decimals, such as 0.29 and 0.80 for 20% and 80% and 0.0773 for kg water, etc. Good work!

Thank you Dr.Bob! In the original problem it was just 35L and not 35.0L. Above you wrote "keeping the 1.8", perhaps you meant 1.18.
I have a habit of not putting zeros in front of decimal numbers, I did not think they were so significant and also, I omitted them here because of space as sometimes it is convenient to have whole calculations on one line.

You "annotate" the problems beautifully, kind of like an experienced chemistry teacher,but surprisingly like a chess player! "In text" annotations help the student learn where he/she went wrong. Perhaps it works best when the student provide a thougt process. It is interesting to note that just at this moment I view you as a skillful positional chess player whose pragmatic style tend to punish opponents who make the slightest of mistakes.

(Capablanca, Petrosian, Karpov are such great positional players)the latters' styles (petrosian and karpov) are not much to my liking as I really admire styles of Kasparov, Fisher (despite his erratic personality; but it's his chess moves that count)both, super aggressive attacking geniuses.
Ok... from chemistry to chess. Perhaps there is a nice link. thx again.

Yes, I meant 1.18 and I didn't catch it when I proofed it. I taught for 36 years before retiring so I have some experience at what students are thinking when they work a problem. It helps all the volunteers on this board for students to show us their work for it allows us to quickly know where the thought process went wrong. I never used zeros before the decimal point until about 10 years ago. Now, however, my eyesight is not so great and the period doesn't show up on the computer. Therefore, I often misread a .35 as 35 but I can't misread 0.35. That zero makes me notice the period. I played a lot of chess in my early days but I was only fair. Good enough to win about 90% of the time but then most people didn't play chess when I was a youngster so few had as much experience as I had. Good luck to you.

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