Em tham khảo nha :
\(\begin{array}{l}
{n_{NaOH}} = \dfrac{{334,5}}{{40}} = 8,3625mol\\
{C_{{M_{NaOH}}}} = \dfrac{{8,3625}}{3} = 2,7875M\\
{m_{{\rm{dd}}NaOH}} = 3000 \times 1,115 = 3345g\\
C{\% _{NaOH}} = \dfrac{{334,5}}{{3345}} \times 100\% = 10\%
\end{array}\)