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I think the dihydrate form is more soluble right? I expect because of the sulphate group attached to two molecules of water.

Also, do you know what's the solubility of each in g/L?

Thank you in advance for any explanation!

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According to the Wikipedia Infobox, solubility products are given::

$4.93 × 10^{−5}\text{ mol}^2\text{L}^{−2}$ (anhydrous)

$3.14 × 10^{−5}$ (dihydrate) [1]

The reference is difficult to trace online, but the theory is sound. In the presence of water anhydrous calcium sulfate spontaneously converts to the dihydrate, so the dihydrate must have lower free energy and therefore less solubility in equilibrium with water. In effect, if you try to dissolve the anhydrous compound into water, the solution becomes oversaturated with respect to the dihydrate and thus the latter precipitates.

Reference

  1. D.R. Linde (ed.) CRC Handbook of Chemistry and Physics, 83rd Edition, CRC Press, 200.
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In mole par liter, $\ce{CaSO4}$ and $\ce{CaSO4·2H2O}$ have the same solubility : $0.018$ mol/L, according to J. Chem. Ed. $77$, Vol. $12$, p. $1558$, Dec. $2000$

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