What makes the enthalpy of hydration of calcium chloride dihydrate unique?
The heats of hydration of your salts, plus some others, can be obtained by subtracting the heat of formation of the anhydrous from the heat of formation of the hydrated salt, using the numerical values from the Handbook of Chemistry and Physics, 43rd Ed., 1961, pages 1807 - 1831. The so obtained values are rather different from yours, but the tendency is the same. Here are the obtained molar heats of hydration, divided by the number of water molecules. They are strangely similar.
$\ce{Na2CO3·H2O....... 295.3 kJ (mol H2O)^{-1}}$
$\ce{Na2CO3·10 H2O..... 295.3 kJ (mol H2O)^{-1}}$
$\ce{NaCH3COO·3H2O ... 297.6 kJ (mol H2O)^{-1}}$
$\ce{Na2SO4·10H2O ..... 294.3 kJ (mol H2O)^{-1}}$
$\ce{Na2S·4.5H2O ...... 305.0 kJ (mol H2O)^{-1}}$
$\ce{Na2S·5H2O....... 304.6 kJ (mol H2O)}^{-1}$
$\ce{Na2S·9H2O ....... 301.1 kJ (mol H2O)^{-1}}$
$\ce{Na2S2O3·5H2O..... 297.2 kJ (mol H2O)^{-1}}$
$\ce{Ca(NO3)2·4H2O.... 297.7 kJ (mol H2O)^{-1}}$
$\ce{Ca(CH3COO)2·H2O . 291.2 kJ (mole H2O)^{-1}}$
$\ce{CaCl2·2H2O ...... 179.6 kJ (mol H2O)^{-1}}$  ........ (!)
$\ce{CaCl2·6 H2O ......301.8 kJ (mol H2O)^{-1}}$
It is immediately clear that the heat of hydration of one molecule $\ce{H2O}$ in these compounds is $\ce{300 ± 10 kJ (mol H2O)^{-1}}$. But there is one odd exception : $\ce{CaCl2·2H2O}$. Has anybody an explanation ?
I would have been happy to show my previous numerical values in a table, or an array with lines and several columns, showing the different heats of formation. But I have not been able to write it correctly. There was always a message saying : \hline incorrect. Can somebody tell me how to proceed ? Thank you in advance.