\(n_{NaOH}=0,25.1=0,25\left(mol\right)\)
PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
Theo PT: \(n_{CH_3COOH}=n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=0,25.60=15\left(g\right)\Rightarrow m_{ddCH_3COOH}=\dfrac{15}{12\%}=125\left(g\right)\)