A gas diffuses twice as fast as CF4 (g). The molecular mass of the gas is:
I'm not sure how to figure this out without the rates of diffusion.
6 answers
The easy way to do it is to make up a number (use any convenient number) for CH4, then make the rate twice as fast for the unknown gas.
I used the formula given and came up with 352 g/mol, would that be correct?
I don't think so. If the unknown diffuses twice as fast it MUST be smaller (thqat is must have a molar mass less than that of CF4)
If I call CF4 rate = 1L/min = r1
Then rate of unknown is 2L/min = r2
molar mass M1 = CF4 = 88
r1/r2 = (sqrt M2/M2)
1/2 = (sqrt M2/88)
Solve that for M2. You first step is to square both side so you can get rid of the sqrt sign.
If I call CF4 rate = 1L/min = r1
Then rate of unknown is 2L/min = r2
molar mass M1 = CF4 = 88
r1/r2 = (sqrt M2/M2)
1/2 = (sqrt M2/88)
Solve that for M2. You first step is to square both side so you can get rid of the sqrt sign.
It makes senses that the molar mass would be smaller than that of CH4 if it diffuses faster. 1/2 = square root of 22/88 = 22 g/mol ?
That's what I obtained, too.
I see you picked up on my typo, I wrote M2/M2 and it should have been sqrt(M2/M1)
I see you picked up on my typo, I wrote M2/M2 and it should have been sqrt(M2/M1)
Yep, I caught it. Thanks for your help!