a, Ta có: mKNO3 = 0,25.101 = 25,25 (g)
\(\Rightarrow C\%_{KNO_3}=\dfrac{25,25}{25,25+175}.100\%\approx12,61\%\)
b, Giả sử: mH2O = x (g)
Có: \(\dfrac{50}{50+x}.100\%=10\%\)
\(\Rightarrow x=450\left(g\right)\)
Vậy: mH2O = 450 (g)
c, Có: \(C\%_{BaCl_2\left(20^oC\right)}=\dfrac{35,8}{35,8+100}.100\%=26,36\%\)
Bạn tham khảo nhé!