00:02
Hi there.
00:03
There are a couple of ways to work this problem.
00:06
Since i do notice that it is at 273 kelvin and one atm, that is standard temperature and pressure.
00:14
So we could use molar volume at stp and go from there.
00:18
But i am going to just go ahead and use the ideal gas law equation.
00:23
So that's pv equals nrt.
00:26
But for moles, we know that moles are equal to the mass.
00:35
Divided by the molar mass.
00:39
Mass divided by molar mass, we'll give you moles.
00:43
So i'm going to substitute that in for n.
00:48
So m is mass.
00:50
I'm using capital m for molar mass, and then times rt.
00:56
Rearranging this to solve for molar mass, i'll multiply both sides by molar mass and divide both sides by pv.
01:05
So what we get is mass times rt divided by pv.
01:14
All right, so we have an equation that we can use just to plug our information in and solve for molar mass.
01:20
So let's go ahead and do that.
01:22
Molar mass is equal to the mass.
01:24
We're told that we have a 1 .00 gram sample.
01:30
Universal gas constant, since we see pressure in atmospheres, i will use the one that is 0 .0 821 liters times atmospheres over mole, times kelvin and then the temperature we are given in the problem as 273 kelvin...