\(m_{ddNaOH}=240\cdot1,1=264\left(g\right)\\ m_{NaOH\left(dd\right)}=264\cdot10\%=26,4\left(g\right)\)
Đặt số mol Na cần thêm là a (mol)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
Mol a a a 0,5a
\(m_{dd\left(spu\right)}=23a+264-a=22a+264\left(g\right)\)
\(C\%_{NaOH\left(spu\right)}=\dfrac{\left(40a+26,4\right)}{22a+264}\cdot100=30\%\\ \Leftrightarrow40a+26,4=6,6a+79,2\\ \Leftrightarrow33,4a=52,8\Rightarrow a=\dfrac{264}{167}\\ \Rightarrow m_{Na}=\dfrac{264}{167}\cdot23\approx36,359\left(g\right)\)