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I thought I understood this, but I'm having some doubts now. My question is quite simple: Can the volume of a solid (specifically a granular crystal or powder) be accurately determined using the displacement method with a liquid in which the solid is soluble?

Originally I thought by dissolving in the liquid there would (or could) somehow be a change to the density of the mixture which would affect the displacement. As such, it would be a good idea to ensure the substance is not soluble in the liquid being used to determine the solid's volume. But if the dissolution of the solid is only a physical process (i.e., dissolving sugar or salt in water), would this at all affect one's ability to measure the volume of the amount of substance added?

I guess when dealing with an unknown substance, we might not be able to easily determine if the dissolution took place as a physical or chemical process without further testing, so it would make more sense to conduct the measurement with a liquid in which the substance was insoluble.

I haven't found a qualified answer to this question online or in my textbook. I'm not being asked this question directly, but I'm incorporating this discussion in my latest lab report on a question related to measuring physical properties.

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  • $\begingroup$ Thanks for the advice! I'll see what others might have to say on the topic. $\endgroup$ – MrCMedlin Sep 21 '14 at 22:04
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Volume is not a conserved extensive property of mixtures in the same way that other quantities like mass, net electric charge, number of particles, etc. are (within closed systems). That is to say, in reality, the volumes of components physically mixed together are not simply additive, with the total volume not necessarily equal to the sum of the individual volumes. I.e., $\sum{V_i} \not= V_{mix}$ generally. This occurs chiefly because intermolecular forces of attraction between particles will vary with the composition of the mixture. Mixtures of ethanol and water, for example, are always reduced in volume by comparison to the summed volumes of ethanol and water, respectively, in isolation.

In fact, the total volume can be calculated given the partial molar volumes of the individual components for a specific composition under given conditions. Specifically, $V_{mix} = \sum{n_i \bar{V_i}}$, where $\bar{V_i}$ and $n_i$ are the partial molar volumes and moles, respectively, of the separate constituents of the mixture.

A few good references I found for additional study: 1, 2. Any decent physical chemistry textbook should also cover partial molar properties.

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  • $\begingroup$ Thank you for this explanation and your references. I've never heard of partial molar properties before. I also didn't direct my search towards volume, as I should have. Does this remain true when the substance does not dissolve in the liquid? Do we have the potential to measure an inaccurate displacement due to the interaction of the outer atoms of our solid and the atoms of the liquid? $\endgroup$ – MrCMedlin Sep 21 '14 at 22:03
  • $\begingroup$ @MrCMedlin, if the substance doesn't dissolve, or the liquid components are largely immiscible, then the two phases remain separated and virtually no interaction occurs between the molecules of the disparate pure substances (except at the surface/phase interface). Consequently, as far as I know, the same phenomenon doesn't occur as it does with homogeneous solutions, and the volumes are simply additive (to within some possible tiny deviation). $\endgroup$ – Greg E. Sep 21 '14 at 22:07
  • $\begingroup$ Yes, I was mostly referring to the contact between the substances at the surface of the solid. I agree this would be negligible and likely near impossible to measure without highly sensitive equipment and/or extremely large amounts of both substances. But thank you, that does answer my second question! $\endgroup$ – MrCMedlin Sep 21 '14 at 22:10
  • $\begingroup$ You have to be careful if you are using a powder that can trap air pockets. $\endgroup$ – MaxW Nov 23 '17 at 23:29

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