c,
$n_{Cl_2}=0,25(mol)$
$n_{NaOH}=0,8.2=1,6(mol)$
$Cl_2+2NaOH\to NaClO+NaCl+H_2O$
$V_{dd}=800ml=0,8l$
Sau phản ứng:
$n_{NaClO}=n_{NaCl}=0,25(mol)$
$\to C_{M_{NaClO}}=C_{M_{NaCl}}=\dfrac{0,25}{0,8}=0,3125M$
$n_{NaOH}=1,6-0,25.2=1,1(mol)$
$\to C_{M_{NaOH}}=\dfrac{1,1}{0,8}=1,375M$