Equivalent Weight And Normality
Hardchemistry
The equivalent weight of KMnO₄ when it acts as an oxidising agent in acidic medium is: (K = 39, Mn = 55, O = 16)
Select the correct option:
Solution
Incorrect! Answer:
31.6
- Identify the Half-Reaction: In acidic medium, MnO4− is reduced to Mn2+:
- MnO4−+8H++5e−→Mn2++4H2O
- Determine n-factor: The number of electrons gained per formula unit = 5 (Mn goes from +7 to +2).
- Calculate Molar Mass of KMnO₄:
- M=39+55+4(16)=39+55+64=158 g/mol.
- Apply the Formula:
- Equivalent weight=n-factorMolar mass=5158=31.6 g/eq.
- Important Distinction: The equivalent weight of KMnO₄ changes with the medium. In neutral medium (n-factor = 3), it would be 52.7, and in strongly basic medium (n-factor = 1), it would be 158. Always identify the medium first.
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About This Question
- Subject
- chemistry
- Chapter
- some basic concepts in chemistry
- Topic
- equivalent weight and normality
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
31.6
- Identify the Half-Reaction: In acidic medium, MnO4− is reduced to Mn2+:
- MnO4−+8H++5e−→Mn2++4H2O
- Determine n-factor: The number of electrons gained per formula unit = 5 (Mn goes from +7 to +2).
- Calculate Molar Mass of KMnO₄:
- M=39+55+4(16)=39+55+64=158 g/mol.
- Apply the Formula:
- Equivalent weight=n-factorMolar mass=5158=31.6 g/eq.
- Important Distinction: The equivalent weight of KMnO₄ changes with the medium. In neutral medium (n-factor = 3), it would be 52.7, and in strongly basic medium (n-factor = 1), it would be 158. Always identify the medium first.
This hard difficulty chemistry question is from the chapter some basic concepts in chemistry, covering the topic of equivalent weight and normality. It appeared in the 2025 exam.
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