\(C1:\\ n_{NaOH}=1.0,1=0,1mol\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,1mol\\ C_{M\left(NaOH\right)}=\dfrac{0,1}{0,1}=1M\\ C2:\\ n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2n_{Na_2O}=0,2mol\\ C_{\%NaOH}=\dfrac{0,2.40}{100+6,2}\cdot100=7,53\%\)