\(n_{Na}=\frac{46}{23}=2\left(mol\right)\)
\(PTHH\text{: }2Na+2H_2O\rightarrow2NaOH+H_2\)
\(TheoPT:\text{ }n_{H_2}=\frac{1}{2}n_{Na}=1\left(mol\right)\Rightarrow m_{H_2}=2.1=2\left(g\right)\)
\(m_{ddspư}=m_{Na}+V_{H_2O}-m_{H_2}=46+224-2=268\left(g\right)\)
\(n_{NaOH}=n_{Na}=2\left(mol\right)\Rightarrow m_{NaOH}=2.40=80\left(g\right)\)
\(C\%_{ddNaOH}=\frac{80}{268}.100\%=29,85\%\)