
#1
Feb2509, 08:18 PM

P: 12

1. The problem statement, all variables and given/known data
A person will receive a DWI if his/her blood alcohol level (BAL) is 100mg per dL of blood or higher. Suppose a person is found to have a BAL of 0.033 mol of ethanol per liter of blood. Will the person recieve a DWI? Assuming the cop isn't cool and just follows you home. 2. Relevant equations molarity and basic metric conversions. 3. The attempt at a solution 100mg/dL = 1000mg/dL = 1g/dL? 0.033 mol of C2H5OH (ethanol) x 46.07g/1 mol = 1.52g/L of C2H5OH Therefore the drunk driver would get a ticket. I think this is correct. I did the problem backwards. What's the best way to get mg/dL from the 0.033 mols of C2H5OH? Thanks for the help :D 



#2
Feb2609, 02:57 AM

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#3
Feb2609, 08:03 AM

P: 12

I meant 100g/dL not 100mg/dL
You're right. 0.033 mol/L. it is in the original question, yes I for got to add that in. now that you have successfully shown me all my typos, can you please answer my original question? What is the correct way of answering this problem to come up with a solution of mg/dL. I can ask my professor if its too much trouble. I'm trying to learn how to be more anal with my calculations. I know it is a big part of chemistry. I have a social science degree so math is my weak point. Thanks for your help. 



#4
Feb2609, 08:36 AM

P: 348

Blood Alcohol Level?how about this, define your conversions more robustly: 1L = 10dL therefore 100mg/dL = 1000mg/L = 1g/L(this is your alcohol limit in terms of Litres) I'll assume your mol calculation is correct (it seems good), then there are 1.52g of Ethanol in 0.033 mol of Ethanol. So the person has 1.52g of Ethanol per litre of blood. This is higher than the 1g/L limit, so he should be arrested for DUI. 



#5
Feb2609, 08:46 AM

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#6
Feb2609, 10:31 AM

P: 12

So to get from 1.52 g/L of C2H5OH to mg/dL 1.52g/L (1000mg/1g per liter) = 1520 mg/L (1520 mg) / (10 dL/ 1L) = 152 mg/dL of C2H5OH The answer in the back of the book is 150 mg/dL so my molar mass calc.s are different from the books, but close enough. I was doing this: 1520 mg/L = (1520mg) (10dL/1L) and getting 15200 mg/dL (I forgot that it is mg PER dL, ha. makes sense now) __I do need to pay closer attention to units/states/detail. Thanks for pointing this out Borek. Criticism helps even though I don't enjoy it. I've made SEVERAL errors on my exams this same way costing me letter grades in some cases . Any good suggestions to avoid this, besides the obvious double checking? I'm slightly A.D.D. 



#7
Feb2609, 10:54 AM

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#8
Feb2609, 11:02 AM

P: 12

Sig. fig.s! ahhh yes. I forgot 150 is only 2 significant figures. tricky.
Thanks again. 



#9
Feb2709, 01:45 AM

P: 348





#10
Feb2709, 12:10 PM

P: 12

VERY helpful, thank you! 



#11
Feb2709, 12:24 PM

P: 12

BAL 100 mg/dL
Person 0.033 mol/L C2H5OH 0.033 mol/L * 46.07g C2H5OH/1 mol (cancel out mols) * 1000 mg/1 g (cancel out g) * 1 L /10 dL (cancel out L) = 152.031....sig. figs....[1.5 x10^2 mg/dL] :D easy. 


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