NTA Abhyas JEE Main2020ChemistryRedox ReactionsPractice
Volume of 0.1 M K 2 C r 2 O 7 required to oxidize 35 mL of 0.5 M F e S O 4 solution is
Options
- A29.2 mL
- B175 mL
- C185 mL
- D145 mL
Correct answer
A. 29.2 mL
Step-by-step solution
C r 2 O 7 2 - + 14 H + + 6 e - → 2 C r 3 + + 7 H 2 O a 1 = 6 a 2 = 1 a 1 = n - factor of K 2 C r 2 O 7 a 2 = n - factor of F e S O 4 F e 2 + → F e 3 + + e - a 1 M 1 V 1 K 2 C r 2 O 7 a 1 = 6 = a 2 M 2 V 2 F e S O 4 a 2 = 1 M 1 = 0.1 M M 2 = 0.5 M V 1 = ? V 2 = 35 mL So V 1 = 1 × 0.5 × 35 mL 6 × 0.1 = 29.2 mL.