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21.08 \times 10^{23} molecules of \mathrm{KMnO_{4}} reacts with excess sulphuric acid to give how many atoms of oxygen?

Option: 1

5.27 \times 10^{23}


Option: 2

52.7 \times 10^{23}


Option: 3

26.35 \times 10^{23}


Option: 4

2.635 \times 10^{23}


Answers (1)

best_answer

21.08\times 10^{23} molecules of \mathrm{KMnO_{4}=3.5} moles of \mathrm{KMnO_{4}}

2 moles of \mathrm{KMnO_{4}} reacts in sulphuric acid to give 5 mole of oxygen atoms.

1 mole of \mathrm{KMnO_{4}} reacts in sulphuric acid to give 5/2 mole of oxygen atoms.

So, 3.5 moles of \mathrm{KMnO_{4}} reacts in sulphuric acid to give 5/2 \times 3.5 = 8.75 moles of oxygen atom = 8.75 \times 6.023 \times 10^{23} = 52.7 \times 10^{23}

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avinash.dongre

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