A bottle of `H_2O_2` is labeled as `15` vol `H_2O_2`. 112 mL of this solution of `H_2O_2`is titrated against `0.05` M ac

Ask a Question


Question Asked by a Student from EXXAMM.com Team


Q 1157612584.     A bottle of `H_2O_2` is labeled as `15` vol `H_2O_2`. 112 mL of this solution of `H_2O_2`is titrated against `0.05` M acidified solution of `KMnO_4`. The volume of `KMnO_4` required is
A

`0.6` L

B

` 1.2` L

C

`2.4` L

D

`12.0` L

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

Find instant solution to any question from our database of 1.5 Lakh + Questions

 


Access free resources including

  • 100% free video lectures with detailed notes and examples
  • Previous Year Papers
  • Mock Tests
  • Practices question categorized in topics and 4 levels with detailed solutions
  • Syllabus & Pattern Analysis

Engineering


Medical


Banking


DEFENCE


SSC


RRB


CBSE-NCERT


OLYMPIAD