a) $2NaOH + Cl_2 \to NaCl + NaCl + H_2O$
b) $n_{Cl_2} = \dfrac{6,5}{22,4} = 0,29(mol)$
$n_{NaOH} = 2n_{Cl_2} = 0,58(mol)$
$V_{dd\ NaOH} = \dfrac{0,58}{1} = 0,58(lít)$
$n_{NaCl} = n_{NaClO} = n_{Cl_2} = 0,29(mol)$
$C_{M_{NaCl}} = C_{M_{NaClO}} = \dfrac{0,29}{0,58} = 0,5M$