$Na_2O +H_2O\xrightarrow{} 2NaOH$

$n_{Na_2O}=\dfrac{31}{23×2+16}=0,5(mol)$

$→n_{NaOH}=0,1(mol)$

$a)C_{M_{NaOH}}=\dfrac{1}{0,5}=2M$

$b) PTHH : 2NaOH +H_2SO_4 \xrightarrow {} Na_2SO_4 +H_2O$

$n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,5(mol)$

$m_{H_2SO_4}=0,5 ×98 =49(g)$

$m_{dd H_2SO_4}=49 ×\dfrac{20}{100}=245(g)$

Cho Khối lượng riếng là $1,14g/ml$

$V_{ddH_2SO_4}=\dfrac{245}{1,14}≈214,9(lít)$