Submit your paper to J Biol Methods today!      # Super easy calculation that I can't figure out what I'm doing wrong

Best Answer Borek, 23 October 2013 - 12:44 PM

1.17 g/mL is not density of HCl, but of the solution. She calculated correctly mass of HCl necessary ((1 L)*(0.5 mol/1L)*(36.46 g/1 mol)) and then converted this mass to teh volume - but using density not of a pure, liquid HCl (which you don't have, and which has a density of 1.5 g/mL when liquid - which is below -85 deg C), but of the 37% solution.

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### #1 knuf

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Posted 23 October 2013 - 12:06 PM

Ok, so a very inquisitive tech asked me a question about a liquid calculation and while I know the right answer, I can't seem to figure out what's wrong with her calculation.

Here is the way I always calculate it. We need 1 L 0.5 N HCl from 12 N (37%) HCl.

ml= (0.5 N)*(1L)  = 41.67 ml + water up to 1 L

12N

Easy right? Ok, so here is the way she looked at it instead using the density of 37% HCl, which is 1.17 g/ml. She did this because the density is written on the bottle, but not the normality.

ml = (1 L)*(0.5 mol/1L)*(36.46 g/1 mol)*(1 ml/1.17 g) = 15.58 ml + water up to 1 L

I've realized that if I invert the density such that it is 1.17 ml/g then it gives me essentially the right answer, but I can't figure out why I would do that. Help me so that I can stop thinking about this stupid equation and get back to wasting time some other way!

### #2 Borek

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Posted 23 October 2013 - 12:44 PM   Best Answer

1.17 g/mL is not density of HCl, but of the solution. She calculated correctly mass of HCl necessary ((1 L)*(0.5 mol/1L)*(36.46 g/1 mol)) and then converted this mass to teh volume - but using density not of a pure, liquid HCl (which you don't have, and which has a density of 1.5 g/mL when liquid - which is below -85 deg C), but of the 37% solution.

### #3 doxorubicin

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Posted 23 October 2013 - 12:51 PM

She also assumed that the HCl was 100% rather than 37%.

### #4 knuf

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Posted 23 October 2013 - 01:37 PM

Of course...yes...density is not concentration. Thank you for your help. 