Giải thích các bước giải:
$n_{Na_2O}=\frac{6,2}{62}=0,1\ \text{mol}$
$Na_2O+H_2O\to 2NaOH$
$\Rightarrow n_{NaOH\ \text{tạo ra}}=0,1.2=0,2\ mol$
$n_{NaOH}=\frac{133,8.44,84}{100.40}=1,5\ \ mol$
$\Rightarrow n_{NaOH}=1,5+0,2=1,7\ \text{mol}$
BTKL: $m_{Na_2O}+m_{dd\ NaOH}=m_{dd\ spu}=6,2+133,8=140 \ gam$
$\Rightarrow \text C\%_{dd}=\frac{1,7.40}{140}\cdot 100\%=48,6\%$