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Question: 100 mL of acidified 2N \({{\text{H}}_{2}}{{\text{O}}_{2}}\) is allowed to react with \(\text{KMn}{{\...

100 mL of acidified 2N H2O2{{\text{H}}_{2}}{{\text{O}}_{2}} is allowed to react with KMnO4\text{KMn}{{\text{O}}_{4}}solution till there is a slight tinge of purple colour. The volume of oxygen produced at STP is:
A. 2.24 L
B. 1.12 L
C. 3.36 L
D. 4.48 L

Explanation

Solution

The volume strength of hydrogen peroxide tells us about the volume of oxygen that is liberated by the one unit of hydrogen peroxide. So, by using volume strength = product of normality and 5.6. Normality is defined as the concentration measurement of the substance, which is defined as the ratio of gram equivalent mass to the per litre of the solution.

Complete Step-by-Step Answer:
-We have to calculate the volume of oxygen which is present in hydrogen peroxide and potassium permanganate.
-The volume produced by both will be the same because according to the Law of multiple proportions if two elements can form more than one compound, then the ratios of the masses of the second element which combine with a fixed mass of the first element will give the whole number ratio.
-Now, we know that the volume of 1N of hydrogen peroxide is equal to the 5.6 volume strength. So, the volume strength of 2N of hydrogen peroxide will be:
× 5.6 = 11.2\text{2 }\times \text{ 5}\text{.6 = 11}\text{.2}
-So, the volume of oxygen produced by 100 mL of 2N hydrogen peroxide will be:
=Volume of H2O2 × Volume strength of H2O2\text{Volume of }{{\text{H}}_{2}}{{\text{O}}_{2}}\text{ }\times \text{ Volume strength of }{{\text{H}}_{2}}{{\text{O}}_{2}}
=100 × 11.2 = 1120 mL100\text{ }\times \text{ 11}\text{.2 = 1120 mL}.
-Now, as stated above that the volume of oxygen produced by hydrogen peroxide and potassium will be the same. So, the volume produced by potassium permanganate will be 1120 mL.
-Hence, the total volume of oxygen produced by the hydrogen peroxide and potassium permanganate at STP (Standard Temperature and Pressure) conditions will be:
1120 + 1120 = 2240 mL1120\text{ + 1120 = 2240 mL} or 22401000 = 2.24 L\text{or }\dfrac{2240}{1000}\text{ = 2}\text{.24 L}
-So, in the given reaction of hydrogen peroxide and potassium permanganate. The volume of oxygen liberated is 2.24L.
2KMnO4 + 5H2O2 + 3H2SO4  2MnSO4 + K2SO4 + 5O2 + 8H2O\text{2KMn}{{\text{O}}_{4}}\text{ + 5}{{\text{H}}_{2}}{{\text{O}}_{2}}\text{ + 3}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{ }\to \text{ 2MnS}{{\text{O}}_{4}}\text{ + }{{\text{K}}_{2}}\text{SO4 + 5}{{\text{O}}_{2}}\text{ + 8}{{\text{H}}_{2}}\text{O}
Therefore, option A is the correct answer.

Note: Volume strength is defined as the volume of oxygen which is liberated by the hydrogen peroxide at the STP conditions. STP conditions have a standard temperature of 298K, standard pressure which is equal to the 1 bar and standard volume 22.4 L.