\(\)\(m_{dd}\left(Na_2CO_3\right)=22,88+77,12=100\left(g\right)\)
\(m_{ct}\left(Na_2CO_3\right)=100.8,48\%=8,48\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{m}{M}=\dfrac{8,48}{23.2+12+16.3}=\dfrac{8,48}{106}\approx0,08\left(mol\right)\)
\(CTHH\left(X\right)=Na_2CO_3.xH_2O\)
\(\Rightarrow n_{\left(Na_2CO_3.xH_2O\right)}=0,08\left(mol\right)\)
\(M_{\left(Na_2CO_3.xH_2O\right)}=\dfrac{m}{n}=\dfrac{22,8}{0,08}=286\left(g/mol\right)\)
\(M_{\left(xH_2O\right)}=286-106=180\left(g/mol\right)\)
\(n_{\left(xH_2O\right)}=\dfrac{m}{M}=\dfrac{180}{18}=10\left(mol\right)\)
Vì 1 mol \(Na_2CO_3.xH_2O\) chứa \(x\) mol \(H_2O\) nên \(x=n_{\left(H_2O\right)}=10\)
\(\Rightarrow x=10\)
Vậy công thức của muối ngậm nước là \(Na_2CO_3.10H_2O\).
