$m_{dd Na_2CO_3}=11,44+88,56=100g$
$\to m_{Na_2CO_3(dd)}=100.4,24\%=4,24g$
$\to n_{Na_2CO_3(\text{tinh thể})}=n_{Na_2CO_3(dd)}=\dfrac{4,24}{106}=0,04(mol)$
$m_{H_2O(\text{tinh thể})}=11,44-4,24=7,2g$
$\to n_{H_2O(\text{tinh thể})}=\dfrac{7,2}{18}=0,4(mol)$
$\to n_{Na_2CO_3}: n_{H_2O}=0,04:0,4=1:10$
Vậy muối ngậm nước là $Na_2CO_3.10H_2O$