The solution would be like this for this specific problem:
4 NaOCl + S2O3{2-} + 2 OH{-} → 2 SO4{2-} + H2O + 4 NaCl
(0.00456 L) x (0.100 mol/L
S2O3{2-}) x (4 mol NaOCl / 1 mol S2O3{2-}) x (100.0 mL / 25 mL) x
(74.4422 g NaClO/mol) = 0.54313 g
(5.00 mL) x (1.08 g/mL) = 5.40 g solution
(0.54313 g) / (5.40 g) = 0.101 = 10.1%
So, the average percent by mass of NaClO in the commercial bleach is 10.1%.