JEE Main202127 Jul 2021Evening ShiftChemistryRedox ReactionsActual
10 . 0 mL of 0 . 05 M KMnO 4 solution was consumed in a titration with 10 . 0 mL of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ........ × 10 - 2 g / L . (Round off to the nearest integer)
Correct answer
0
Step-by-step solution
n eq KMnO 4 = n eq H 2 C 2 O 4 · 2 H 2 O or, 10 × 0 . 05 1000 × 5 = 10 × M 1000 × 2 ∴ Conc. of oxalic acid solution = 0 . 125 M = 0 . 125 × 126   g / L = 15 . 75   g / L = 1575 × 10 - 2   g / L