\(m_{ct}=\dfrac{4.100}{100}=4\left(g\right)\)
\(n_{NaOH}=\dfrac{4}{40}=0,1\left(mol\right)\)
Pt : \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,1 0,1
\(n_{HCl}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
20ml = 0,02l
\(C_{M_{ddHCl}}=\dfrac{0,1}{0,02}=5\left(M\right)\)
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