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CBSE Class 12-science Answered

1.What is the theory behind the titration of kmno4 solution against mohr salt? what is the change in the mass of mohr salt to be used if  M/10 molar solution of mohr salt is taken instead of M/20 solution?
2. the following question is not related to the above question. can we draw a graph taking the following scale
X axis:1unit= 0.5 ampere
y axis:1 unit = 0.2 volt.
if the readings are correct ,will we obtain a straight line for the above scale &calculate its slope correctly?  the reading is based on ohms law
 
Asked by Ajnu C S | 01 Feb, 2015, 12:14: AM
answered-by-expert Expert Answer
Dear drshajiajnu@gmail.com

Thanks for asking us a question in Ask the Expert section of TopperLearning.com.

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Answer to first question:

The reaction between KMnO4 and Mohr’s salt is a redox reaction and the titration is therefore called a redox titration. Mohr’s salt is the reducing agent and KMnO4 is the oxidizing agent. KMnO4 acts as an oxidizing agent in all the mediums; i.e. acidic, basic and neutral medium. KMnO4 acts as the strongest oxidizing agent in the acidic medium and therefore dil. H2SO4 is added to the conical flask before starting the titration. KMnO4 solution is deep violet in colour due to presence of MnO- 4. In presence of dilute H2SO4, it reacts with the reducing agent (Ferrous sulphate in Mohr’s salt) and gets reduced to Mn2+ which is colourless ions, hence the pink colour disappears. This process continues till all the reducing agent has been oxidised, then at this stage the excess drop of KMnO4 added is not reduced and pink colour is observed in the solution.

Molecular Mass of Mohr’s Salt = 392 g/mole

1 L of 1M KMnO4 requires 392g Mohr’s Salt.

M/20 KMnO4 requires 392/20 g or 19.6 g Mohr’s salt.

M/10 KMnO4 requires 392/10 g or 39.2 g Mohr’s salt.

 
Regards

Topperlearning Team.
Answered by Arvind Diwale | 02 Feb, 2015, 12:12: PM
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