Ta có:$n_{K_2O}=0,2 mol$$K_2O + H_2O\rightarrow 2KOH$$m_{KOH}=22,4 g$$C$%$=\frac{22,4.100}{m_{dd}}=5,6\rightarrow m_{dd}=\frac{22,4.100}{5,6}=400 (g)$Mà: $m_{dd}=m_{ct}+m_{H_2O}=22,4+m_{H_2O}=400\rightarrow m_{H_2O}=377,6(g)$
Ta có:$n_{K_2O}=0,2 mol$$K_2O + H_2O\rightarrow 2KOH$$m_{KOH}=22,4 g$$C$%$=\frac{22,4.100}{m_{dd}}=5,6\rightarrow m_{dd}=\frac{22,4.100}{5,6}=400 (g)$Mà: $m_{dd}=m_{ct}+m_{H_2O}=18,8+m_{H_2O}=400\rightarrow m_{H_2O}=381,2(g)$
Ta có:$n_{K_2O}=0,2 mol$$K_2O + H_2O\rightarrow 2KOH$$m_{KOH}=22,4 g$$C$%$=\frac{22,4.100}{m_{dd}}=5,6\rightarrow m_{dd}=\frac{22,4.100}{5,6}=400 (g)$Mà: $m_{dd}=m_{ct}+m_{H_2O}=
22,
4+m_{H_2O}=400\rightarrow m_{H_2O}=3
77,
6(g)$