Đáp án + Giải thích các bước giải:

$m_{NaOH}=400.8\%=32(g)$

$a\big)$

$m_{dd\,sau}=400+200=600(g)$

$\to C\%_{dd\,sau}=\dfrac{32}{600}.100\%\approx 5,33\%$

$b\big)$

$C\%_{dd\,sau}=\dfrac{32}{300}.100\%\approx 10,67\%$

$c\big)$

$m_{dd\,NaOH\,12\%}=200.12\%=24(g)$

$\to C\%_{dd\,sau}=\dfrac{24+32}{400+200}.100\%\approx 9,33\%$

$d\big)$

$n_{Na_2O}=\dfrac{6,2}{62}=0,1(mol)$

$Na_2O+H_2O\to 2NaOH$

Theo PT: $n_{NaOH}=2n_{Na_2O}=0,2(mol)$

$\to m_{NaOH}=0,2.40=8(g)$

$\to C\%_{dd\,sau}=\dfrac{8+32}{400}.100\%=10\%$

$e\big)$

$C\%_{dd\,sau}=\dfrac{32+10}{400}.100\%=8,25\%$

$m_{NaOH}$ =$\frac{400 .8}{100}$ =32(g)

a,  $m_{dd}$ =400+200=600(g)

C%dd=$\frac{32}{600}$ .100%=5,33%

b,   C%dd=$\frac{32}{300}$ .100%=10,67%

c,    $Na_{2}O$ +$H_{2}O$ →2NaOH

$n_{Na_{2}O}$ =$\frac{6,2}{62}$ =0,1(mol)

Theo PT:$n_{NaOH}$ =2$n_{Na_{2}O}$=0,2(mol)

⇒$m_{NaOH}$=0,2 . 40=8(g)

C%dd=$\frac{8}{400}$ .100%=2%

d, $m_{NaOH}$=10+32=42(g)

C%dd=$\frac{42}{400}$ .100%=10,5%

Cho mình 5⭐và 1 lượt cám ơn và câu trả lời hay nhất nhé!