nHCl = 0,1 mol
a/ [HCl] = 0,2M
b/ V= 0,1 : 0,1 = 1 lit
=> cần thêm 500ml nước
a) \(n_{HCl}=\dfrac{3,65}{36,5}=0,1\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)
b) \(V_{ddHCl.0,1M}=\dfrac{0,1}{0,1}=1\left(l\right)=1000\left(ml\right)\)
\(\Rightarrow V_{H_2O}thêm=1000-500=500\left(ml\right)\)