\(n_{Na}=\dfrac{1,15}{23}=0,05\left(mol\right)\)
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,05------------>0,05------>0,025
\(\rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,05.40+120.15\%=20\left(g\right)\\m_{dd}=1,15+120-0,025.2=121,1\left(g\right)\end{matrix}\right.\\ \rightarrow C\%_{NaOH}=\dfrac{20}{121,1}.100\%=16,12\%\)