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$$C{O_2}$$  is obtained only when the compound is decomposed by acid thus, it is $$N{a_2}C{O_3}.$$   It cannot be $$NaHC{O_3}$$   since, it gives $$C{O_2}$$  on heating.
Probable compound is $$\mathop {N{a_2}C{O_3}}\limits_{85.48\% } \cdot \mathop {x{H_2}O}\limits_{14.52\% } $$

\[\begin{align} & \begin{matrix} {} & \text{Percentag} & \text{Molar ratio} & \text{Ratio} \\ N{{a}_{2}}C{{O}_{3}} & 85.48 & 0.8064 & 1 \\ {{H}_{2}}O & 14.52 & 0.8066 & 1 \\ {} & {} & {} & {} \\ \end{matrix} \\ & \text{Thus, compound is}\,N{{a}_{2}}C{{O}_{3}}\cdot {{H}_{2}}O \\ & \text{Molecular weight}=124\,g\,mo{{l}^{-1}} \\ & \text{Percentage of sodium} \\ & =\frac{2\times 23}{124}\times 100 \\ \end{align}\]
  $$ = 37.1\% $$

$$MgS{O_4}$$  is the only alkaline earth metal sulphate which is soluble in water and for solubility hydration energy should be greater than lattice energy.

Calcium carbonate on thermal decomposition gives $$CaO$$  (Basic oxide) and $$C{O_2}$$  (Acidic oxide)
\[CaC{{O}_{3}}\xrightarrow{\Delta }\underset{\text{Basic oxide }}{\mathop{CaO}}\,+\underset{\text{Acidic oxide }}{\mathop{C{{O}_{2}}\uparrow }}\,\]

$$N{a^ + }$$ ion has larger size than $$M{g^{2 + }}$$  ion and hence hydration energy of $$M{g^{2 + }}$$  is larger than that of $$N{a^ + }$$.

Washing soda is chemicaly named as sodium carbonate decahydrate, so its formula is $$N{a_2}C{O_3} \cdot 10{H_2}O.$$

As the size of the anion increases, the covalent character increases and hence the $$m.p.$$  decrease.

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