\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ KL:A\left(I\right)\\ A+H_2O\rightarrow AOH+\dfrac{1}{2}H_2\\ n_A=n_{AOH}=0,1.2=0,2\left(mol\right)\\ M_A=\dfrac{4,6}{0,2}=23\left(\dfrac{g}{mol}\right)\\\Rightarrow A:Natri\left(Na=23\right)\\ m_{NaOH}=0,2.40=8\left(g\right)\\ m_{ddNaOH}=4,6+60-0,1.2=64,4\left(g\right)\\ C\%_{ddNaOH}=\dfrac{8}{64,4}.100\approx12,42\%\\ \Rightarrow B\)