$\text{Bài 1:}$
$n_{N_2O}=\dfrac{6,2}{62}=0,1(mol)$
$\text{Phương trình:}$
$Na_2O+H_2O\to 2NaOH$
$\to n_{NaOH}=2n_{Na_2O}=0,2(mol)$
$\to m_{NaOH}=0,2.40=8g$
$\text{Do ta có:} m_{\text{dd}}=6,2193,8=200g$
$\to \text{C%}_{NaOH}=\dfrac{\text{8.100%}}{200}=4$%
$\text{Bài 2:}$
$m_{\text{dd}}=m_{NaCl}+m_{H_2O}=16+34=50g$
$\to \text{C%}=\dfrac{\text{16.100%}}{50}=32$%