I understand the answer to the question "Is it possible to have Volume(solution) < Volume(solvent) + Volume(solute)?" is "yes" -- but it is possible that adding a solute causes the solvent to contract so much that the volume of the solution is less than the initial volume of the solvent?

An ideal answer would include either a specific example of such a case, or an explanation of why it is impossible.

If the above is not possible, is it possible for the volume of the solution to equal the initial volume of the solvent?

  • $\begingroup$ @Zhe I would count that only if the total volume of the mixture were less than the initial volume of one of the components (either the water or the alcohol, but not the sum of the two). $\endgroup$
    – zacronos
    Oct 6, 2017 at 13:45
  • $\begingroup$ I remember encountering this question elsewhere. The expert opinion was that there is no law forbidding this, but a convincing example remained elusive. $\endgroup$ Oct 6, 2017 at 14:12
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    $\begingroup$ @IvanNeretin I would accept that as an answer if you can locate and cite that expert opinion. $\endgroup$
    – zacronos
    Oct 6, 2017 at 14:28
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    $\begingroup$ I will not post this as an answer because it does not meet my own standards. The question is really good, it deserves a more comprehensive handling . $\endgroup$ Oct 6, 2017 at 14:39
  • $\begingroup$ Sorry, I misread your question. I'm not sure how this contraction is going to take place. Liquids are generally incompressible since there is not much intermolecular space. So what you're asking is essentially to compress an essentially incompressible material and add more matter into that mix... $\endgroup$
    – Zhe
    Oct 6, 2017 at 15:00

1 Answer 1


Gosh, sure. According to various on-line info (of unknown quality) a 2% by wt. solution of NaOH in water has a density of 1.0207 at 20°C. This means 100 g of it has a volume of 97.97 ml. At 20°C water has a density of 0.9982 meaning that 98 grams of water has a volume of 98.18 ml.


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