Đáp án + Giải thích các bước giải:

$a\big)$

$n_{SO_2}=\frac{4,48}{22,4}=0,2(mol)$

$SO_2+H_2O\leftrightharpoons H_2SO_3$

Theo PT: $n_{H_2SO_3}=n_{SO_2}=0,2(mol)$

Ta có $m_{dd\,spu}=0,2.64+147,2=160(g)$

$\to C\%_{H_2SO_3}=\frac{0,2.82}{160}.100\%=10,25\%$

Ta có $V_{dd\,H_2SO_3}=\frac{160}{1,05}\approx 152,4(ml)=0,1524(l)$

$\to C_{M\,H_2SO_3}=\frac{0,2}{0,1524}\approx 1,31M$

$b\big)$

$n_{Na_2O}=\frac{15,5}{62}=0,25(mol)$

$Na_2O+H_2O\to 2NaOH$

Theo PT: $n_{NaOH}=2n_{Na_2O}=0,5(mol)$

$\to C\%_{NaOH}=\frac{0,5.40}{134,5+15,5}.100\%\approx 13,33\%$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
S{O_2} + {H_2}O \to {H_2}S{O_3}\\
{n_{S{O_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2mol\\
{m_{S{O_2}}} = 0,2 \times 64 = 12,8g\\
{n_{{H_2}S{O_3}}} = {n_{S{O_2}}} = 0,2mol\\
{m_{{H_2}S{O_3}}} = 0,2 \times 82 = 16,4g\\
C{\% _{{H_2}S{O_3}}} = \dfrac{{16,4}}{{0,2 \times 64 + 147,2}} \times 100\%  = 10,25\% \\
{V_{{H_2}S{O_3}}} = \dfrac{{160}}{{1,05}} = 152ml = 0,152l\\
{C_{{M_{{H_2}S{O_3}}}}} = \dfrac{{0,2}}{{0,152}} = 1,32M\\
b)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{15,5}}{{62}} = 0,25mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,5mol\\
{m_{NaOH}} = 0,5 \times 40 = 20g\\
C{\% _{NaOH}} = \dfrac{{20}}{{15,5 + 134,5}} \times 100\%  = 13,33\% 
\end{array}\)