Gases

[The Cancer Cure]

1. A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 67 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 24 s for 1.0 L of O2 gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

 

Find g/mol.

 

Don't know what the heck to do.

 

2. A gaseous mixture of O2 and Kr has a density of 1.110 g/L at 425 torr and 300. K. What is the mole percent O2 in the mixture?

 

Find %.

 

Quote answer from number 1.

 

Specific thanks to herman.

[hermanntrude]

OK question 1 required Grahame's law, which can be written like this:

 

a ratio of any of the following properties for a pair of gases is equal to the square root of the ratio of their molar masses:

 

• molecular speeds
  
• effusion rates
  
• effusion times
  
• distances travelled by molecules
  
• amounts of gases effused
  

 

the best way to use this law is to reason it out quantitatively first, ask yourself "should the ratio be greater or less than one?" In this particular question, that can be rephrased as "would you expect the unknown gas to have a higher or lower molecular weight than oxygen based on the fact it effused more slowly?" then use the ratio of molar masses whichever way up works for your reasoning.

 

then just use the times and molecular weights to make an equation, plug in the numbers and solve for the molecular weight of the unknown.

[The Cancer Cure]

The ratio of the unknown gas to the oxygen gas should be greater.

 

So...I'm slightly confused.

 

24^2 + 32 (Oxygen Gas) = 67^2 + X (Unknown)

 

? Maybe ?

[hermanntrude]

The ratio of the unknown gas to the oxygen gas should be greater.

 

?

 

what I think you mean is that because the effusion time was longer, the mass of the unknown gas should be greater than that of oxygen.

 

Perhaps I didn't explain grahame's law very well. In this case it looks like this, but bear in mind the ratio of molar masses might be the other way up for some of the listed properties above:

 

[math]\frac{{effusion time O_{2}}}{{effusion time unknown gas}}[/math]=[math]\sqrt{\frac{M_{O_{2}}}{M_{unknown}}}[/math]

 

I think that's how it works, anyway. I got myself a bit confused with the Latex stuff and now i've lost track of my thoughts

[The Cancer Cure]

Hm, so:

 

24/67 = square root of 32/x

 

so:

 

(24/67)^2 = 32/x

 

.1283136556 = 32/x

 

x = 249.3888889

 

With the sig figs, x = 250.

 

And that fine sir, is the correct answer.

 

Thank you so much!

 

 

What about the last question? - Last one for the night.

[hermanntrude]

Hm, so:

 

24/67 = square root of 32/x

 

so:

 

(24/67)^2 = 32/x

 

.1283136556 = 32/x

 

x = 249.3888889

 

With the sig figs, x = 250.

 

And that fine sir, is the correct answer.

 

Thank you so much!

 

 

What about the last question? - Last one for the night.

 

OK i helped you quite a lot with that one. I'd just like to remind you that you have to construct your own equations in this type of question so make sure you can see how i put that together. it may be worth reading the section in your textbook on grahame's law so you consolidate the knowledge.

 

As for question 2, try solving the ideal gas law (in the form where you can use density as a variable) for M. This M will be the average M of all the atoms involved. from that you should be able to solve for the mole fraction by rearranging the equation for an average.

 

I'm going to bed now so good luck with this one. good night!

[The Cancer Cure]

Thanks for all of your help. I appreciate it!

 

'night.