PTHH: 2KOH + H2SO4 ---> K2SO4 + 2H2O
Theo PT: \(n_{KOH}=2.n_{H_2SO_4}=2.0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{KOH}=0,4.56=22,4\left(g\right)\)
Ta có: \(C_{\%_{KOH}}=\dfrac{22,4}{m_{dd_{KOH}}}.100\%=10\%\)
\(\Rightarrow m_{dd_{KOH}}=224\left(g\right)\)
Theo đề, ta có: \(D=\dfrac{224}{V_{dd_{KOH}}}=1,12\left(\dfrac{g}{ml}\right)\)
\(\Rightarrow V_{dd_{KOH}}=200\left(ml\right)\)