a, PTHH : \(NaOH+HCl\rightarrow NaCl+H_2O\)
\(n_{HCl}=C_M.V=0,2.1=0,2\left(mol\right)\)
- Theo PTHH : \(n_{NaOH}=n_{HCl}=0,2\left(mol\right)\)
=> \(V_{NaOH}=\frac{n}{C_M}=\frac{0,2}{1,5}=\frac{2}{15}\left(l\right)\)
b, - Theo PTHH : \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
=> \(m_{NaCl}=n.M=0,2.58,5=11,7\left(g\right)\)
Ta có : \(V_{dd}=0,2+\frac{2}{15}=\frac{1}{3}\left(l\right)\)
=> \(C_M=\frac{n}{V}=\frac{0,2}{\frac{1}{3}}=0,6\left(M\right)\)