Question

How many grams of $KMn{{O}_{4}}$will react with 50ml of 0.2M ${{H}_{2}}{{C}_{2}}{{O}_{4}}$solution in the presence of ${{H}_{2}}S{{O}_{4}}$?
(A) 1.58g
(B) 3.16g
(C) 0.632g
(D) 0.79g

Answer 1

Hint Molarity of any given solution can be defined as the total number of moles of solute per litre solution. It is dependent upon the physical properties of the solution. The mass of $KMn{{O}_{4}}$ can be obtained by using the formula of Molarity shown below
$Molarity=\dfrac{no.of moles}{volume}$.

Complete step by step answer:
The chemical reaction for the question be written as-
$2KMn{{O}_{4}}+5{{H}_{2}}{{C}_{2}}{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+10C{{O}_{2}}$
There are few chemical reactions very sensitive to the exact amount of reagent to be used otherwise the reaction does not work properly. In this reaction, once the exact concentration has been determined, the solution is standardized, and then the exact amount of $KMn{{O}_{4}}$can be added to the chemical reaction.
Given values-
Volume of ${{H}_{2}}S{{O}_{4}}$= 50ml
i.e. 0.05lts
Molarity, M = 0.02M
Since, $Molarity=\dfrac{no.of moles}{volume}$
After substituting the given values, we get
$0.2=\dfrac{no.of moles}{0.05}$
No. of moles = 0.01
$2KMn{{O}_{4}}+5{{H}_{2}}{{C}_{2}}{{O}_{4}}+3{{H}_{2}}S{{O}_{4}}\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+8{{H}_{2}}O+10C{{O}_{2}}$
By this reaction,
2 moles of $KMn{{O}_{4}}$= 5 moles of ${{H}_{2}}{{C}_{2}}{{O}_{4}}$
Therefore, moles of KMn{{O}_{4}}$$$$=\dfrac{2}{5}x0.01 = 0.004
Molecular weight of $KMn{{O}_{4}}$= 158g/mol
$n=\dfrac{mass}{Molarmass}$
$0.004=\dfrac{mass}{158}$
$mass=0.632g$
Therefore, Mass of $KMn{{O}_{4}}$is 0.632g

So, the answer to the above question is (C) 0.632g

Additional information:
Potassium permanganate is a powerful oxidizing agent. It shows many used in organic chemistry. It is also used as a medication for dermatitis, for cleaning wounds, and general disinfection.

Note: $KMn{{O}_{4}}$ serves as self-indicator in acidic solution. Potassium permanganate is standardized against pure ${{H}_{2}}{{C}_{2}}{{O}_{4}}$, oxalic acid. It involves a redox reaction. Oxalic acid is oxidised to carbon dioxide by $KMn{{O}_{4}}$ which itself gets reduced to $MnS{{O}_{4}}$. In order to keep the reaction environment acidic, we put the acid in Erlenmeyer flask and potassium permanganate in the burette.