\(K_2O+H_2O\rightarrow2KOH\)
0,05-------------> 0,1
\(n_{K_2O}=\dfrac{4,7}{94}=0,05\left(mol\right)\)
\(C\%_{X\left(KOH\right)}=\dfrac{0,1.56.100\%}{4,7+195,3}=2,8\%\)
\(C\%=\dfrac{m_{CT}}{m_{dd}}.100\%\\ =\dfrac{4,7}{195,3}.100\%\approx2,4\%\)
`N_(K_2O)= (4,7)/94=0,05mol`
PƯ : `K_2O+H_2O -> 2KOH`
`=> C% (ddKOH)=(0,1 *56)/(4,7 +195,3) * 100%=2,8%`
KL dung dịch=KL chất tan+KL dung môi
=4,7+195,3=200 g
\(C\%=\dfrac{m_{CT}}{m_{dd}}.100\%\\ =\dfrac{4,7}{200}.100\%=2,35\%\)
