# Does solubility matter when measuring density with displacement method?

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.

• Thanks for the advice! I'll see what others might have to say on the topic. – MrCMedlin Sep 21 '14 at 22:04

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.