Rates of Effusion: Homework Questions & Solutions

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SUMMARY

The discussion focuses on the rates of effusion of gases, specifically addressing three homework problems involving the calculation of molar masses and effusion rates. The first problem calculates the molar mass of an unknown gas based on its effusion rate relative to nitrogen, yielding a result of 131 g/mol. The second problem analyzes the effusion rates of uranium isotopes UF6, resulting in a ratio of 0.9934. The third problem identifies krypton as the unknown noble gas by correctly applying the effusion formula, demonstrating the importance of maintaining consistent variable positioning in calculations.

PREREQUISITES
  • Understanding Graham's Law of Effusion
  • Familiarity with molar mass calculations
  • Knowledge of isotopes and their separation techniques
  • Basic proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study Graham's Law of Effusion in detail
  • Learn about the properties and applications of UF6 in isotope separation
  • Explore the concept of effusion time and its relationship to gas properties
  • Practice solving problems involving molar mass and effusion rates
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Chemistry students, educators, and professionals involved in gas behavior studies, particularly those focusing on effusion and molar mass calculations.

Agent M27
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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 separated 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
 
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You have just got momentarily confused into thinking the bigger number meant faster, but as it is time taken the bigger number means slower, so in your you first attempt you put it upside down. The ratio of atomic masses is close to 4 and Kr is clearly the answer they want.

(You might object that "effuses from a container in 76 seconds" is a pretty vague statement, but whatever it is it is the same for both and Kr is the answer they want. :biggrin:)
 

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