\(n_{BaO}=\dfrac{15.3}{153}=0.1\left(mol\right)\)
\(n_{H_2O}=\dfrac{1.8}{18}=0.1\left(mol\right)\)
\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)
\(0.1.......0.1...........0.1\)
=> BaO tan hết
\(V_{dd}=\dfrac{1.8}{1}=1.8\left(ml\right)=1.8\cdot10^{-3}\left(l\right)\)
\(C_{M_{Ba\left(OH\right)_2}}=\dfrac{0.1}{1.8\cdot10^{-3}}=55.5\left(M\right)\)