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  • 500 mL of 56 V H2O2 is kept in an open container due to which some H2</

500 mL of 56 V H2O2 is kept in an open container due to which some H2O2 is decomposed and evolves 8gm O2 simultaneously, during the process some H2O also vapourizes. Due to all these changes final volume is reduced by 20%. Find final volume strength of H2O2 (aq).

Option: 1

56


Option: 2

44.8


Option: 3

11.2


Option: 4

33.6


Answers (1)

best_answer
Vol strength of solution = 56 V

Therefore, Vol of O2 produced by 500ml solution = 56*500ml = 28 L

Now, Moles of O2 produced = 28/22.4 = 1.25 mol

The chemical equation is given as follows:

2H_{2}O_{2}\, \rightarrow \, O_{2}\, +\, 2H_{2}O 

Therefore from the stoichiometry calculations, we have:
1 mole of O2 will be produced by 2 moles H2O2 

Therefore, 1.25 moles of O2 will be produced by 2.5 mol of H2O2

So, mass of H2O2 = 2.5*34= 85g

Now, the density of H2O2 = 170g/L

Again, as given in the question, the amount of O2 evolved = 8g or 0.25 mol

Hence 0.5 mol of H2O2 decomposes to produce 0.5 mol of H2O and  0.25 mol of O2

Loss = 20% of 85 g  = mass of O2 escaped + mass of H2O vapourised

=> mass of H2O vapourised = 14.45 - 8 = 6.45 g

So final solution contains only H2O2

Mass of final solution = 85 - 14.45 = 70.55 g

Final vol = 70.55/170 = 415 ml

68g H2O2 gives 32g O2

=> 415 ml of solution gives 1.04 moles of O2 or 23296 ml O2

=> 1ml solution gives = 23296/415 = 56.1 ml of O2

Hence vol strength = 56.1 V

Therefore, Option (1) is correct

Posted by

Deependra Verma

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