Question

1.)A 1.998-g sample containing Cl' and CIO," was dissolved in sufficient water to give 250.0 mL of solution. A 50.00-ml aliquot required 13.97 mL of 0.08551 M AgNO3 to titrate the CI'. A second 50.00-ml aliquot was treated with V2(SO4)3 to reduce the CIOA to Cl": CIO:+4V2(SO)) + 4H20 C1 + 1250,2 +8VO2 + 8H Titration of the reduced sample required 40.12 mL of the AgNO, solution. Calculate the percentages of Cl' and CIO, in the sample.

Answer 1

This simple titration reaction follows the simple 1:1 stoichiometry and first measurement of Cl^- is done applying the formula of molarity then measurement of ClO₄^- is done using the concept that it reduces to give the Cl^- which is in terms of mole both are same. this way both the compound and their percentage can be found out.
chemical Reaction ci + Agt at Agel (s) Now, for first so ml solution I using Max der = Magalog & Vagales Не је Ich is V & molanity volume of a solution McLX 5o= D.O8551X 13.97 Plc 50 = 1.195 Mal = 1.195 = 0.023gM. 50 so, in año ml solution Number of roles of cl = Holanity x volume lin liter) = 0.0239 x anco = 5.975 x 103 mole I Mons of ct Molour mass & No. of Moles & 5.975 x 108 x 35.5 1.a12 gm -

chemical Reaction elog - +42(SO4)₂ & 4160 et tlason 2- to voet torrt As clos deduces to the some amount af et so, after to seduction the amount or for Number of modes of ct decides the Number of Aoles o elog i Doing some an pluevious for second 50 ml solution ! Mer- & Veri MANO X V Agoloz. Her y 50 ml = 0.0856 x 40.12 Mar y 50 A 8,431 Me co 31431 20.06862

so, in ano ml Number of Moles of clor - Molarity x volume in liter? 0,06562 Y ano 1000 = 0.0172 Mods of clot in 250 ml solution Moles (Number ) X Molay moros = 0.01727 99.5 = 1.712 gy Son in 1.998 gm sample Amount of ela = 0.212 Amount of clot = 1.712 af of el- = 0.22 1.998 0.12 2100 X100 0. 1061X 100 10.611 % of cl04_ 1412 X 100 1998 08966. 0.85.69 YIOD 85.69 )