Question

If the molecular weight of $KMn\mathop O\nolimits_4 $ is 158 ,then the equivalent weight (nearest integer) is___________ in acidic medium.

Answer 1

Hint: Equivalent weight is defined as weight of one equivalent of substance and is shown by M/n where M is molar mass of substance and n is the n-factors which is defined as acidity or basicity of a substance or electrons gained in the reaction .

Complete step by step answer:
As when $KMnO_4$ is added to water it becomes an acidic medium. The reaction is as follows:-
$Mn\mathop O\nolimits_4^ - + 8\mathop H\nolimits^ + + 5\mathop e\nolimits^\_ \to M\mathop n\nolimits_2^ + + 4\mathop H\nolimits_2 O$
 As 5 electrons are gained in the reaction so its n factor becomes 5 . Now we know that the Molecular weight of $KMnO_4$ is 158 ,so we're going to put the value of M and n
Now the equivalent is $\dfrac{M}{n}$ . It can be written as $\dfrac{M}{5}$
So it will become $\dfrac{{158}}{5} = 31.6$ rounding off it will become 32.

Note:
-$KmN\mathop O\nolimits_4 $ is also known as potassium permanganate.
-It has dark purple crystals.
-It has intense colour and weak temperature dependent diamagnetism.
-Pi bonding takes place by overlapping the p orbital of oxygen with d orbital of manganese.
-It is a good oxidising agent in acidic, basic or neutral medium.
-It is prepared by fusion of $Mn\mathop O\nolimits_2 $ with an alkali metal hydroxide and an oxidising agent like $KNO_3$.
-Its salt is not very much soluble in water.