Đặt \(n_{Na_2CO_3.xH_2O}=a\left(mol\right)\)
\(\Rightarrow n_{Na_2CO_3}=a\left(mol\right)\)\(\Rightarrow m_{Na_2CO_3}=106a\left(g\right)\)
Theo đề, ta có: \(4,24=\dfrac{106a.100}{5,72+44,82}\)
\(\Rightarrow a=0,02\left(mol\right)\)
\(\Rightarrow x=10\)