\(n_{Na}=\dfrac{23}{23}=1\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(1......................1.............0.5\)
\(m_{NaOH}=500\cdot1.2\cdot10\%=60\left(g\right)\)
\(m_{NaOH}=1\cdot40+60=100\left(g\right)\)
\(m_{dd_{NaOH}}=23+500\cdot1.2-0.5\cdot2=622\left(g\right)\)
\(C\%_{NaOH}=\dfrac{100}{622}\cdot100\%=16.07\%\)