## Chemistry and Chemical Reactivity (9th Edition)

$40.05\ mg$
Number of moles of $Na_2S_2O_3$ consumed: $0.0425\ M\times 25.3\ mL=1.08\ mmol$ From stoichiometry: $1.08\ mmol\ S_2O_3^{2-}\times\dfrac{1\ mmol\ I_2}{2\ mmol\ S_2O_3^{2-}}\times\dfrac{2\ mmol\ I^-}{1\ mmol\ I_2}\times\dfrac{1\ mmol\ HClO}{2\ mmol\ I^-}= 0.538\ mmol\ HClO$ Which is the number of moles of sodium hypochlorite in the sample, so the mass is: $0.538\ mmol\times 74.44\ mg/mmol=40.05\ mg$