\(m_{CH_3COOH}=100.6\%=6\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)
PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)
Theo PT: \(n_{NaHCO_3}=n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow a=C_{M_{NaHCO_3}}=\dfrac{0,1}{0,05}=2\left(M\right)\)