Dissolve 0,1256g powder (it contains chloramine) in 100 mL water.
Add 1 g of potassium iodide R and 5 mL of dilute sulfuric acid R and leave it for 3 minutes.
Titrate with 0.1 M sodium thiosulfate using 1 mL starch solution R as an indicator.
8,44 mL was titrated
f = 1,02123
ClO- + 3 I- + 2 H+ -> Cl- + I3- + H2O
I3- + 2 S2O3^2- -> 3 I- + S4O6^2-
Calculate mass percent (%m/m) of anhydrous chloramine
mass chloramine = [(M x f x V)sodium thiosulfate x Mr anhydrous chloramine]/2
= (0,1mM x 1,02123 x 8,41mL x 227,64mg/mmol)/2
= 195,5096 mg
mass percent = ( 195,5096 mg / 125.6 mg ) x 100 % = 156%
Where did i go wrong? the mass percent should be above 50% and below 100%