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.
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 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.
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)
Yep, I caught it. Thanks for your help!
To determine the molecular mass of the gas, we can use Graham’s law of diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
Let's denote the rate of diffusion of the unknown gas as r1 and the rate of diffusion of CF4 gas as r2. According to the problem, the unknown gas diffuses twice as fast as CF4, so we can set up the following equation:
r1 = 2 * r2
Now, Graham’s law of diffusion states that the ratio of the rates of diffusion is equal to the square root of the ratio of the molar masses:
r1/r2 = √(M2/M1)
Since we are comparing the unknown gas (M1) to CF4 (M2), we can rewrite the equation as:
2 * r2 / r2 = √(M2 / M1)
Canceling out the r2 terms, we have:
2 = √(M2 / M1)
To solve for M1 (molecular mass of the unknown gas), we need to isolate it in the equation. Squaring both sides, we get:
4 = M2 / M1
Multiplying both sides by M1, we have:
4M1 = M2
Therefore, the molecular mass of the unknown gas is four times the molecular mass of CF4 (M2). To find the molecular mass of CF4, you can refer to the periodic table and calculate the sum of the atomic masses of carbon and four fluorine atoms.