Ban đầu: $m_{NaOH}=m.10\%=0,1m(g)$
$n_{Na}=\dfrac{9,2}{23}=0,4 mol$
$2Na+2H_2O\to 2NaOH+H_2$
$\Rightarrow n_{H_2}=0,2 mol$
$\Sigma m_{NaOH}=0,1m+0,4.40=0,1m+16(g)$
$m_{dd}=9,2+m-0,2.2=m+8,8(g)$
$C\%=17,56\%$
$\Rightarrow \dfrac{0,1m+16}{m+8,8}=0,1756$
$\Leftrightarrow m=191,2$