A/Hoà tan 20g NaOH vào 100g H2O

⇒C% NaOH=$\frac{20}{20+100}$.100≈16,667 %

B/Hoà tan 6,2g Na2O vào 100g H2O

n Na2O=$\frac{6,2}{62}$=0,1 mol

Na2O+H2O→2NaOH

 0,1 →              0,2              mol

mdd sau pứ=m Na2O+m H2O

                    =6,2+100  =106,2 g

⇒C% NaOH=$\frac{0,2.40}{106,2 }$.100≈7,533 %

C/Hoà tan 4,6g Na vào 100g H2O

n Na=$\frac{4,6}{23}$=0,2 mol

2Na+2H2O→2NaOH+H2↑

0,2→                0,2        0,1          mol

mdd sau pứ=m Na+m H2O-m H2

                    = 4,6+100-0,1.2 =104,4 g

⇒C% NaOH=$\frac{0,2.40}{104,4 }$.100≈7,66 %

------------------------Nguyễn Hoạt----------------------------

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}NaOH}} = {m_{NaOH}} + {m_{{H_2}O}} = 20 + 100 = 120g\\
{C_\% }(NaOH) = \dfrac{{20}}{{120}} \times 100\%  = 16,67\% \\
b)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{6,2}}{{62}} = 0,1\,mol \Rightarrow {n_{NaOH}} = 0,2\,mol\\
{m_{{\rm{dd}}}} = 6,2 + 100 = 106,2g\\
{C_\% }(NaOH) = \dfrac{{0,2 \times 40}}{{106,2}} \times 100\%  = 7,53\% \\
c)\\
{n_{Na}} = \dfrac{{4,6}}{{23}} = 0,2\,mol\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{NaOH}} = {n_{Na}} = 0,2\,mol\\
{n_{{H_2}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_{{\rm{dd}}}} = 4,6 + 100 - 0,1 \times 2 = 104,4g\\
{C_\% }(NaOH) = \dfrac{{0,2 \times 40}}{{104,4}} \times 100\%  = 7,66\% 
\end{array}\)