00:01
In this video, we are going to apply gram salt of effusion to calculate the molar mass of an unknown gas.
00:06
We know that the rate of effusion of some arbitrary gas a, so the rate of effusion of some arbitrary gas b, is inversely proportional to the square root of their molar masses.
00:16
So it's the molar mass of b on top and the molar mass of a on the bottom.
00:21
We also know that rate is the amount of gas that effuses per unit of time.
00:29
So if we keep the amount constant, then another way, way that we can write this is the time it takes for b to a fuse.
00:36
I'm flipping everything because it's in the denominator's.
00:38
Over the time that it takes for a to a fuse, and then we keep the other side the same as the molar mass of b over the molar mass of a, all square rooted.
00:48
All right, so we have our unknown, and we have oxygen.
00:53
Let's let oxygen equal gas a, and then our unknown will be gas b.
01:00
That way we'll have our unknown on the top, because it's easier to solve for something on the top than it is on the bottom.
01:06
Okay.
01:06
So the time that it takes the unknown to diffuse is 43 seconds.
01:10
We have the amount the same.
01:11
That's why i'm using this second equation...