Bài 1:
\(m_{NaOH}=\dfrac{150.10}{100}=15\left(g\right)\)
mdd (sau) = 150 + 50 = 200 (g)
=> \(C\%_{dd.sau}=\dfrac{15}{200}.100\%=7,5\%\)
Bài 2:
\(m_{CuSO_4\left(bd\right)}=\dfrac{150.10}{100}=15\left(g\right)\)
=> \(m_{CuSO_4\left(sau\right)}=15+20=35\left(g\right)\)
mdd(sau) = 20 + 150 = 170 (g)
=> \(C\%_{dd.sau}=\dfrac{35}{170}.100\%=20,588\%\)


