a. \(n_{NaOH\left(5\%\right)}=\frac{100.5}{100.40}=0,125mol\)
\(m_{ddNaOH\left(8\%\right)}=\frac{0,125.100.40}{8}=62,5g\)
\(m_{H_2O}\) bay hơi \(=100-62,5=37,5g\)
b. \(m_{NaOH\left(A\right)}=40.0,125=5g\)
\(m_{NaOH\left(8\%\right)}=\frac{100.8}{100}=8g\)
\(m_{NaOH}\) thêm vào \(=8-5=3g\)