\(n_{NaOH}=\frac{12,4}{40}=0,31\left(mol\right)\)
=> CM = \(\frac{0,31}{0,5}=0,62M\)
Ta có: \(n_{NaOH}=\frac{12,4}{40}=0,31\left(mol\right)\)
\(\Rightarrow C_{M\left(A\right)}=C_{M\left(NaOH\right)}=\frac{0,31}{0,5}=0,62M\)
Bạn tham khảo nhé!
\(n_{NaOH}=\frac{12,4}{40}=0,31\left(mol\right)\\ C_{M_A}=C_{M_{NaOH}}=\frac{0,31}{0,5}=0,62\left(M\right)\)