\(m_{KOH\left(bđ\right)}=120.5,6\%=6,72\left(g\right)\)
\(n_{K_2O}=\dfrac{0,94}{94}=0,01\left(mol\right)\)
\(m_{dd\left(spư\right)}=0,94+120=120,94\left(g\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
0,01-------------->0,02
\(\rightarrow m_{KOH\left(spư\right)}=0,02.56+6,72=7,84\left(g\right)\\ \rightarrow C\%_{KOH\left(sau\right)}=\dfrac{7,84}{120,94}.100=6,48\%\)
\(C\%=\dfrac{0,94}{120}.100\%=0,783\%\)