# Rates of Effusion

## Homework Statement

So this isn't just one problem but it is a question as to why it is done one way for two instances and then the opposite way for a third problem. Here it goes.

1) An unknown gas effuses at a rate that is .462 times that of nitrogen gas (at the same temperature). Calculate the molar mass of the unknown gas in g/mol.

2) Uranium-235 can be seperated from U-238 by flourinating the uranium to form UF6 (which is a gas) and then taking advantage of the different effusion rates for compounds containing the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6.

3) A sample of neon effuses from a container in 76 seconds. The same amount of an unknown noble gas requires 155 seconds. Identify the gas.

## Homework Equations

$$\frac{RateA}{RateB}$$ = $$\sqrt{\frac{M_B}{M_A}}$$

Where M = molar mass

## The Attempt at a Solution

1) $$\frac{RateA}{RateB}$$ = .462

MA = $$\frac{M_B}{(.462)^2}$$

MA = $$\frac{28.02}{(.462)^2}$$ = 131g

2) Rate A/ MA = U-238 RateB/ MB = U-235

$$\frac{RateA}{RateB}$$ = $$\sqrt{\frac{M_B}{M_A}}$$

$$\sqrt{\frac{235.054}{238.051}}$$= .9934

3) This is the one where I am confused on why it was carried out in the way as follows:
If I was to follow the format as shown above, where rateA is over rateB and then on the other side they switch positions, I arrive at a terribly incorrect answer:

$$\frac{76s}{155s}$$ = $$\sqrt{\frac{M_U}{20.18g}}$$

.2404=$$\frac{M_U}{20.18}$$$$\Rightarrow$$ 4.85g which is the incorrect answer.

The only way I can arrive at the correct answer is when rate A is in the same postion on both sides for example:

$$\frac{76s}{155s}$$ = $$\sqrt{\frac{20.18g}{M_U}}$$

MU = $$\frac{20.18g}{.2404}$$$$\Rightarrow$$ 83.9g

Which points to Krypton and that happens to be the correct answer. I realize this is a long question and I have the answer, but I want to make sure I am not missing something in the formatting and why it appears to change between problems. Thanks in advance.

Joe