JEE Main202228 Jun 2022Morning ShiftChemistryRedox ReactionsActual
A 2 . 0 g sample containing MnO 2 is treated with HCl liberating Cl 2 . The Cl 2 gas is passed into a solution of KI and 60 . 0 mL of 0 . 1 MNaS 2 O 3 is required to titrate the liberated iodine. The percentage of MnO 2 in the sample is____. Nearest integer) [Atomic masses (in u ) Mn = 55 ; Cl = 35 . 5 : O = 16 , I = 127 , Na = 23 , K = 39 , S = 32 ]
Correct answer
0
Step-by-step solution
MnO 2 + 4 HCl ⟶ MnCl 2 + Cl 2 g + 2 H 2 O Cl 2 + 2 Kl ⟶ I 2 + 2 KCl I 2 + 2 Na 2   S 2 O 3 ⟶ 2 Nal + Na 2   S 4 O 6 Mili eq. of MnO 2 = Mili eq. of Cl 2 = Mili eq. of I 2 = Mili eq. of Hypo. 2 w 87 = 0 . 1 × 60 W = 261 miligram %   of   MnO 2 = 0 . 261 2 × 100 = 13 . 05 % ≈ 13