\(n_{NaCl}=\dfrac{35,1}{58,5}=0,6\left(mol\right)\)
Pt: \(2NaCl+H_2SO_4\rightarrow Na_2SO_4+2HCl\)
\(\Rightarrow n_{HCl}=n_{NaCl}=0,6\left(mol\right)\)
\(\Sigma_{m_{dd\left(spu\right)}}=m_{HCl}+m_{H_2O}=0,6.36,5+78,1.1=100\left(g\right)\)
\(C\%_{HCl}=\dfrac{0,6.36,5}{100}.100=21,9\%\)
\(V_{HCl}=\dfrac{100}{12}=\dfrac{25}{3}\left(l\right)\)
\(C_{M_{HCl}}=\dfrac{0,6}{\dfrac{25}{3}}=0,072M\)
Câu b giải cũng tương tự thôi, tính m ra -> số mol..