chem help, ideal gas law please :]?

Question

if .2125 grams of a gas is in a container of 1.25 lter volume at a pressure of .838 atm. and 40c, what is the molecular weight of the gas.


i gonverted 40c to K =313K
but i've filled in all of my variables. . . what am i doing wrong
please help!

Answer 1

To begin, the ideal gas law is PV = nRT.
If you are trying to solve for the molecular weight of the gas, then we must solve for n for the number of moles which yields:

n = (PV/RT)

If you plug in all your equations (including the gas constant, R, which is 0.08205), you get:

n = 0.0407 moles.

To find molecular weight, we take the number of grams and divide it by the moles:

(0.2125 grams)/(0.0407 moles)

in which we get our answer of:
5.21 grams/mole

Answer 2

using pV = nRT provides you with n, as long as you exchange each and every of the parts properly, and use the main suitable fee of R. Molar mass = mass/n, and you may desire to then have the skill to % out the gas.

Answer 3

You solve PV=nRT for n, the number of moles.
Now you can form a proportion:
(n moles/0.2125 g) = (1 mole/ mol wt)

Answer 4

dont forget that:
n = grams of substance/molar mass

so n would equal
.2125/molar mass where molar mass is your unknown.

just plug and chug

Answer 5

use o.o82 for r
http://en.wikipedia.org/wiki/Gas_constant