# BR Gen Chem Ch. 1 Passage 4 #23 & 24

Discussion in 'MCAT Study Question Q&A' started by chocposit, May 12, 2014.

1. ### chocposit

So I'm really confused here! I got C, but the answer is B and I would like to know where they got the 58g/mol and well, how they got the answer in general. The explanation is not helping... help.

And if you can also help with #24, because I'm not quite sure why (1.55/22.4) is used... please and thank you!

Last edited: May 12, 2014
3. ### BerkReviewTeachCompany Rep & Bad SingerExhibitor 10+ Year Member

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The 58 grams per mole came from the second experiment. The vapor occupies 1.00 L and weighs 2.32 g. From the table we know that one mole of that gas should occupy around 24.96 L at 304 K (the boiling point and thus vapor temperature). 1/24.96 is roughly 1/25 = 0.04 moles. The MW is 2.32/0.04 g/mole = 2.32 x 25 = about 58 g/mole.

For question #24, look at the units. 1.55 is the L of CO2 collected and 22.41 is L/mole for an ideal gas. So, 1.55/22.41 would have units of L over L/mole = mole. That value represents the moles of CO2 collected in the experiment.

4. ### chocposit

Thank you so much! But can you explain to me how the compact version of question 23 was solved in that way?

Why divide 82.9 by 12 and 17.1 by 1? Why not divide by the MM of C + H = 13g? I feel like I'm missing something here...

5. ### chocposit

I'm still confused... someone shed some light here, please and thank you ~

6. ### chocposit

I figured it out. You divide by each element's amu, not the MM. Wow, forgetting the basics much (confused it with mass %). Never miiiiiind.