Question

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.

Answer 1

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.

Answer 2

I used the formula given and came up with 352 g/mol, would that be correct?

Answer 3

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.

Answer 4

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 ?

Answer 5

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)

Answer 6

Yep, I caught it. Thanks for your help!