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I am trying to solve the following elementary problem of volume percent, however, I am not quite sure where my mistake is:

Ethanol has a volume-by-volume concentration of 93.3%. Calculate the volume of water in $\pu{10 L}$ of this alcohol.

My Attempts:

By definition,

$$\text{volume concentration} = \frac{\text{volume of solute}}{\text{volume of solution}} \times 100\%$$

From the exercise, we know that $\text{volume concentration} =93.3\%$ and $\text{volume of solute} = \pu{10 L}$. By substituting those values in the equation above, we then have

$$93.3\%=\frac{\pu{10 L}} {\text{volume of solution}} \times 100\%$$

Working forward from the last equation and solving it for the volume of the solution, we find

$\text{volume of solution} = \pu{10.71 L}$. Therefore, the volume of water or the volume of solvent becomes

$$\text{volume of solution} = \text{volume of solute} + \text{volume of solvent}$$

$$ \pu{10.71 L} = \pu{10 L} + \text{volume of solvent}$$

Therefore, $\text{volume of solvent} = \pu{0.71 L}$.

However, the actual answer to this problem is $\pu{667 g}$. Based on the above, what is my mistake here?

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  • $\begingroup$ For future reference: for the body of questions, answers and comments, chemistry.se allows to use mhchem. In addition to assistance typing chemical formulae, the \pu{} command allows to type values with units including a non-breakable space (and units typically are upright, too). $\endgroup$
    – Buttonwood
    Commented Aug 22, 2023 at 7:44
  • $\begingroup$ Chem+Math Expression formatting reference: MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Aug 22, 2023 at 7:54

1 Answer 1

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First of all it is a stupid problem.

  • Volumes don't add. So 90.00 ml of ethanol and 10.00 ml of water doesn't yield 100.00 ml of solution.
  • The problem messes up significant figures. 10 L and 93.3% ethanol in the problem statement suddenly become 10.000 liters and 93.33% in the problem solution.
  • The problem statement is given in terms of volumes, but the answer for water is given in grams.

So the problem assumes that volumes do add and that the final solution (a mixture of ethanol and water) has a volume of 10.000 liters. Since the solution is 93.3% ethanol, 6.7% is water. So about 670 ml of water. Now further assuming that water is 1 gram per ml (since no temperature is given) that means that there is about 670 grams of water.


Your mistake - In your second equation the 10 L should be in the denominator not the numerator.

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    $\begingroup$ Well, does "10 L of the alcohol" mean 10 L of the impure liquid, or the volume of liquid that contains 10 L of pure alcohol after separation? I agree with "stupid problem". $\endgroup$
    – Karsten
    Commented Aug 22, 2023 at 16:16
  • $\begingroup$ @Karsten - LOL In order to get the "right" answer it must be 10 L of the ethanol/water mixture. // Getting crotchety in my old age. I think problems out to adhere to the real science. The problems are supposed to be teaching chemistry too. $\endgroup$
    – MaxW
    Commented Aug 22, 2023 at 17:53
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    $\begingroup$ Yes, they ought to. $\endgroup$
    – Karsten
    Commented Aug 22, 2023 at 18:17

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