`a)`

`n_{Na_2O}={6,2}/{62}=0,1(mol)`

`Na_2O+H_2O->2NaOH`

`n_{NaOH}=2n_{Na_2O}=0,2(mol)`

`C\%_{NaOH}={0,2.40}/{6,2+93,8}.100\%=8\%`

`b)`

`100g\ NaOH` có `0,2mol`

`->50g\ NaOH` có `0,1mol`

`2NaOH+H_2SO_4->Na_2SO_4+2H_2O`

`n_{H_2SO_4}=1/2n_{NaOH}=0,05(mol)`

`m=m_{dd\ H_2SO_4}={0,05.98}/{4,9\%}=100(g)`

`c)`

`n_{Na_2SO_4}=n_{H_2SO_4}=0,05(mol)`

`C\%_{Na_2SO_4}={0,05.142}/{50+100}.100\%\approx 4,73\%`

$\rm a)$ 

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

Phản ứng xảy ra:

$\rm\quad\quad\;\;\;Na_2O+H_2O\longrightarrow 2NaOH\\(mol)\:\:\:\:\:0,1\hspace{1,3cm}\to\hspace{1,1cm}0,2$

$\rm m_{NaOH}=0,2×40=8\ (gam)$

$\rm m_{dd\ NaOH}=6,2+93,8=100\ (gam)$

$\rm ⇒C\%_{NaOH}=\dfrac{8}{100}×100\%=8\%$

$\rm b)$

Trong 50 gam dung dịch có 0,1 mol NaOH.

Phản ứng xảy ra:

$\rm\quad\quad\;\;\;2NaOH+H_2SO_4\longrightarrow Na_2SO_4+2H_2O\\(mol)\;\;\;\;\;\;0,1\:\to\:\:\:0,05\hspace{1,6cm}0,05$

$\rm ⇒m_{dd\ H_2SO_4}=\dfrac{0,05×98}{4,9\%}=100\ (gam)$

$\rm c)$

$\rm m_{Na_2SO_4}=0,05×142=7,1\ (gam)$

$\rm m_{dd\ Na_2SO_4}=50+100=150\ (gam)$

$\rm ⇒C\%_{Na_2SO_4}=\dfrac{7,1}{150}×100\%=4,73\%$