\(\ \begin{array}{l}m_{NaOH}=\text{24%.200=48(g)}\\n_{NaOH}=\dfrac{48}{40}=1,2(mol) \\\Rightarrow n_{Na}=1,2(mol) \\\Rightarrow x=m_{Na}=1,2.23=27,6(g) \\n_{H_2}=\dfrac{1}{2}n_{Na}=0,6(mol) \\\Rightarrow m_{dd}=m_{Na}+m_{H_2O}-m_{H_2} \\\Rightarrow y=m_{H_2O}=200-27,6+0,6=173(g)\end{array}\)