Đáp án:

 $m = 191,2(gam)$

Giải thích các bước giải:

có $n_{Na} = \dfrac{9,2}{23} = 0,4(mol)$
$2Na + 2H_2O → 2NaOH + H_2$
Ta có :

$n_{NaOH} = n_{Na} = 0,4(mol)$
$n_{H_2}= \dfrac{n_{Na}}{2} = 0,2(mol)$

Sau khi hòa tan hết :

$m_{NaOH} = 0,4.40 + m.10\% = 16 + 0,1m(gam)$

$m_{dd} = m_{Na} + m_{\text{dd NaOH}} - m_{H_2} = 9,2 + m - 0,2.2 = 8,8+m(gam)$
$⇒ C\%_{NaOH} = \dfrac{16+0,1m}{8,8+m}.100\% = 17,56\%$
$⇒ m = 191,2(gam)$

Ban đầu: $m_{NaOH}=m.10\%=0,1m(g)$

$n_{Na}=\dfrac{9,2}{23}=0,4 mol$ 

$2Na+2H_2O\to 2NaOH+H_2$

$\Rightarrow n_{H_2}=0,2 mol$

$\Sigma m_{NaOH}=0,1m+0,4.40=0,1m+16(g)$

$m_{dd}=9,2+m-0,2.2=m+8,8(g)$

$C\%=17,56\%$

$\Rightarrow \dfrac{0,1m+16}{m+8,8}=0,1756$

$\Leftrightarrow m=191,2$