Imagine I have a bottle of 100ml which weight 90,135gr. I know that there is only water and alcohol in it. I know that the weight of water is 0,998 gr/cm3 and alcohol is 0,79 gr/cm3. And when I try to figure out how it is divided I'm doubting whether it is at all possible or it is by trying out around 42% alcohol. But is that true and how can you easily calculate this?


closed as off-topic by Todd Minehardt, Jon Custer, ringo, bon, M.A.R. May 21 '16 at 6:17

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  • 2
    $\begingroup$ If this is a homework question the answer may be fairly simple. If it isn't you might want to make into account that water and ethanol are not ideal liquids and the volume of the mixture will not equal the volume of the two components added up. $\endgroup$ – matt_black May 20 '16 at 20:10
  • $\begingroup$ It is not home work $\endgroup$ – Marijn May 20 '16 at 20:59

100 $cm^3 =90.135 g$

density is $0.90135 g / cm^3$

$0.90135 g / cm^3$ = $0.998x grams + 0.79y grams/ 100 cm^3$



redefine it as

$0.90135 g / cm^3 = (0.998x grams + 0.79(100-x) grams)/100$

$90.135=(0.998x grams + 0.79(100-x) grams)$

$90.135=0.998x grams + 79 - 0.79x grams$

$90.135=79 - 0.208x grams$

$11.135=0.208x grams$



$53.5336538 mL$ of water.

  • 3
    $\begingroup$ For homework questions we have a policy where you do not simply post the answer without explanation but rather that you give hints that lead the asker towards their own solution. We are not a homework completion service. Also, this is horrible to read. Please format it using LaTeX. $\endgroup$ – bon May 20 '16 at 19:48
  • $\begingroup$ Sorry I messed up on formatting and I removed the answer to the question. $\endgroup$ – Arvin Singh May 20 '16 at 20:07
  • 1
    $\begingroup$ I would contest that the answer is incorrect as volumes of solvent mixtures are not simply additive, nor are they simply linear relationships $\endgroup$ – Beerhunter May 21 '16 at 8:45

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