joonkimdds

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what is the molar mass of a 2L sample of gas that weighs 8 g at a temperature of 15 degree celcius and a pressure of 1.5 atm?

what i did was

22.4L (1atm/1.5atm)(273K/288K) = 15.8L =new volume of that specific set up

D= mass/volume
so the new volume * density = molar mass
and so that is 15.8 * 4 = 63.2 is the molar mass and that's the answer.

Then here is my question.
Why is density = 4?
the solution says D = 8g /2L at 15 celcius and 1.5 atm.
but shouldn't it be 8g / 15.8L?
15.8 L is the volume that I found in that temperature and pressure.
 
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joonkimdds

joonkimdds

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allstardentist said:
no... the volume of 1 mole of gas at STP is 22.4 L
?
 
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thehipster

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This is how I'd go about to solve the eqn, I'm not sure if it's right though.

D=m/V = 8g/2L = 4 g/L

Plug this into the eqn: MW = DRT/P = (4 x 0.082 x 288)/1.5 = 62.98 g/mol

I think this should be the right answer. I'm not sure why you're using 22.4L in your eqn. 22.4L = 1mol of gas under STP conditions only, which this isn't.
 
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joonkimdds

joonkimdds

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thehipster said:
This is how I'd go about to solve the eqn, I'm not sure if it's right though.

D=m/V = 8g/2L = 4 g/L

Plug this into the eqn: MW = DRT/P = (4 x 0.082 x 288)/1.5 = 62.98 g/mol

I think this should be the right answer. I'm not sure why you're using 22.4L in your eqn. 22.4L = 1mol of gas under STP conditions only, which this isn't.

is that how u r supposed to solve this type of unknown gas problem? always? because my Kaplan blue book doesn't show anything about Molar weight = DRT/P even though it seems very nice method after seeing how u solve it.

The kaplan blue book showed the solution but i don't understand at all and here it is.

d=8g/2L at 15 C and 1.5 ATM
Vstp = 2L(273K/288K)*(1.5atm/1atm)= 2.84L
8g/2.84L = 2.82g/L at STP
(2.82g/L)(22.4L/mol ) = 63.2 g/mol

do u understand this method too? or do u recommand me to use ur method?
by the way, this is from kaplan blue book page 264.
 

thehipster

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Kaplan Blue book shows 2 ways of solving for density. On page 263 it shows how D = m/V = (P x MW)/(RT)

You can use the way you were trying to describe, but the way I showed is easier to remember. If you know this relationship above, you should be all set for any problems involving density.

You can probably remember D = m/V pretty easily. A mnemonic for remembering MW = dRT/P is MW = DiRT over Pee. You can still use your method for solving the problem, you just set it up wrong.
 
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