Bài 3:
Ta có: mNaOH (A) = 500.10% = 50 (g)
a, \(C\%=\dfrac{50}{500+100}.100\%\approx8,33\%\)
b, \(C\%_B=\dfrac{50+5}{500+5}.100\%\approx10,89\%\)
c, \(C\%_C=\dfrac{50}{400}.100\%=12,5\%\)
d, mNaOH (thêm vào) = 300.15% = 45 (g)
\(\Rightarrow C\%_D=\dfrac{50+45}{500+300}.100\%=11,875\%\)