The hydrated salt Na₂CO₃.xH₂O undergoes 630/o loss in mass on heating and becomes anhydrous. The value of x is:

${\mathrm{Na}}_2{\mathrm{CO}}_3\cdot {\mathrm{xH}}_2\mathrm{O}\left(\mathrm{s}\right)⟶{\mathrm{Na}}_2{\mathrm{CO}}_3\left(\mathrm{s}\right) +{\mathrm{xH}}_2\mathrm{O}\left(\mathrm{g}\right)$
Let 100 g of Na₂CO₃.xH₂O be present
$\therefore \frac{100\mathrm{x}}{106+18\mathrm{x}}$ mole of H₂O formed
$\begin{array}{l}63=\frac{100\mathrm{x}}{106+18\mathrm{x}}\left(18\right)\\ 6678+1134\mathrm{x}=1800\mathrm{x}\\ 666\mathrm{x}=6678\\ \mathrm{x}\simeq 10\end{array}$