a,
$n_{SO_2}=\dfrac{1,12}{22,4}=0,05(mol)$
- TH1: vừa đủ tạo $NaHSO_3$
$NaOH+SO_2\to NaHSO_3$
$\Rightarrow n_{NaOH}=0,05(mol)$
$C_{M_{NaOH}}=\dfrac{0,05}{0,1}=0,5M$
- TH2: vừa đủ tạo $Na_2SO_3$
$2NaOH+SO_2\to Na_2SO_3+H_2O$
$\Rightarrow n_{NaOH}=0,05.2=0,1(mol)$
$C_{M_{NaOH}}=\dfrac{0,1}{0,1}=1M$
Vậy $0,5\le C_{M_{NaOH}}<1$
b,
$n_{Na}=\dfrac{2,3}{23}=0,1(mol)$
$2Na+2H_2O\to 2NaOH+H_2$
$\Rightarrow n_{NaOH}=0,1(mol); n_{H_2}=0,05(mol)$
$m_{dd\text{spu}}=2,3+100-0,05.2=102,2g$
$\to C\%_{NaOH}=\dfrac{0,1.40.100}{102,2}=3,9\%$