Đáp án:

\(\begin{array}{l}
a)C{M_{KOH}} = 1M\\
b){V_{{\rm{dd}}{H_2}S{O_4}}} = 0,1l
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
{K_2}O + {H_2}O \to 2KOH\\
a)\\
{n_{{K_2}O}} = 0,25mol\\
 \to {n_{KOH}} = 2{n_{{K_2}O}} = 0,5mol\\
 \to C{M_{KOH}} = \dfrac{{0,5}}{{0,5}} = 1M
\end{array}\)

\(\begin{array}{l}
b)\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{1}{2}{n_{KOH}} = 0,25mol\\
 \to {m_{{H_2}S{O_4}}} = 24,5g\\
 \to {m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{24,5}}{{20\% }} \times 100\%  = 122,5g\\
 \to {V_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{122,5}}{{1,14}} = 107,46ml = 0,1l
\end{array}\)

Bạn tham khảo nha!

`-` `K_2O + H_2O -> 2KOH`

a. `-` `500` `(ml)` `= 0,5` `(l)`

`-` $n_{K_2O}$ `=` $\dfrac{23,3}{94}$ `= 0,24` `(mol)` 

`-` Theo phương trình $n_{KOH}$ `= 0,48` `(mol)` 

`->` `C_M =` $\dfrac{n}{V}$ `=` $\dfrac{0,48}{0,5}$ `= 0,96` `(M)` 

b. `-` `2KOH + H_2SO_4 -> K_2SO_4 + 2H_2O`

`-` Theo phương trình $n_{H_2SO_4}$ `= 0,24` `(mol)` 

`->` $m_{H_2SO_4}$ `= 0,24 × 98 = 23,52` `(g)`

`->` $m_{dd}$ `=` `\frac{23,52}{20%} × 100% = 117,6` `(g)` 

`->` $V_{dd}$ `=` $\dfrac{117,6}{1,14}$ `= 103,15` `(ml)` `= 0,15` `(l)`