Use app×
Ask a Question
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE

5.5 g of a mixture of FeSO4 .7H2O and Fe2(SO4)3 . 9H2O required 5.4 mL of 0.1 N KMnO4

← Prev Question Next Question →
+1 vote
135 views
in Chemistry by (55.9k points)
closed by

5.5 g of a mixture of FeSO4 .7H2O and Fe2(SO4)3 . 9H2O required 5.4 mL of 0.1 N KMnO4 solution for complete oxidation. Calculate the moles of hydrated ferric sulphate in the mixture.

Play Quiz Game >

1 Answer

+1 vote
by (54.4k points)
selected by
 
Best answer

Mn7+O4- → Mn+2 (charge transfered 5 electrons)

Therefore weight of Fe2(SO4)3 · 9H2O = 5.5 - 0.150 g = 5.350 g

Moles of Fe2(SO4)3 · 9H2O = 5.350 / 562 

= 9.5 x 10–3 mole

← Prev Question Next Question →

Find MCQs & Mock Test

  1. JEE Main 2025 Test Series
  2. NEET Test Series
  3. Class 12 Chapterwise MCQ Test
  4. Class 11 Chapterwise Practice Test
  5. Class 10 Chapterwise MCQ Test
  6. Class 9 Chapterwise MCQ Test
  7. Class 8 Chapterwise MCQ Test
  8. Class 7 Chapterwise MCQ Test

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

User Contribution to Sarthaks eConnect
Managed by Sarthaks Digitizing India
...