\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)
\(m_{dd}=a+b\left(g\right)\)
`Na_2O + H_2O -> 2NaOH`
Theo PT: `n_{NaOH} = 2n_{Na_2O} = a/(31) (mol)`
\(\Rightarrow C\%_{NaOH}=\dfrac{40.\dfrac{a}{31}}{a+b}.100\%=10\%\)
\(\Leftrightarrow\dfrac{40a}{31}=0,1\left(a+b\right)\\ \Leftrightarrow\dfrac{369a}{310}=0,1b\\ \Leftrightarrow\dfrac{a}{b}=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}\)
Hay `a : b = 31 : 369`
\(Na_2O+H_2O->2NaOH\\ C\%=0,1=\dfrac{\dfrac{a}{62}\cdot2\cdot40}{a+b}\\ 0,1a+0,1b=\dfrac{40}{31}a\\ 0,1b=\dfrac{369}{310}a\\ a:b=\dfrac{0,1}{\dfrac{369}{310}}=\dfrac{31}{369}=0,08401\)
