$n_{KMnO_4}=\dfrac{6,32}{158}=0,04(mol)$
$2KMnO_4+16HCl\to 2KCl+2MnCl_2+5Cl_2+8H_2O$
$\to n_{Cl_2}=\dfrac{5}{2}n_{KMnO_4}=0,1(mol)$
$n_{NaOH}=0,15(mol)$
$Cl_2+2NaOH\to NaCl+NaClO+H_2O$
$\Rightarrow Cl_2$ dư, $NaOH$ hết
$n_{NaCl}=n_{NaClO}=\dfrac{n_{NaOH}}{2}=0,075(mol)$
$\to C_{M_{NaCl}}=C_{M_{NaClO}}=\dfrac{0,075}{0,15}=0,5M$