To 70 mg of a sample containing nitrobenzene, 35.00 ml of 0.1000 M TiCl3 solution was added.

To 70 mg of a sample containing nitrobenzene, 35.00 ml of 0.1000 M TiCl3 solution was added. The excess of the reagent was titrated with a 0.0500 M solution of the Fe3 + salt with ammonium thiocyanate as an indicator; for titration went 10.00 ml. What is the nitrobenzene content in the sample? Round your answer to the nearest whole number.

C6H5NO2 + 6 TiCl3 + 6 HCL = C6H5NH2 + 6TiCl4 + 2H2O
TiCl3 + FeCl3 = TiCl4 + FeCl2
So, first, let’s find the quantity of each reagent:
TiCl3: 0.035 * 0.1 = 0.0035 mol
FeCl3: 0.01 * 0.05 = 0.0005 mol
quantity for determination of nitrobenzene
0.0035-0.0005 = 0.003 mol
amount of nitrobenzene 0.003 / 6 = 0.0005 mol
mass of nitrobenzene = 0.0005 * 123 grams / mol = 0.0615 grams
mass fraction = 0.0615 gram / 0.07 gram * 100% = 87.86% = 88%
Answer: 88%