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Question

Find the number of electrons present in a pure sample of MgSO4.7H2O containing 5.5 moles of oxygen atoms?
(Take Avogadro's number (NA)=6×1023)

A
3.9×1025
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B
4.2×1023
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C
5.8×1025
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D
4.5×1023
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Solution

The correct option is A 3.9×1025
From the formula unit of (MgSO4.7H2O), 11 moles of O atoms are present in 1 mole of MgSO4.7H2O.

Therefore, 5.5 moles of O atoms will be present in
=5.511 moles of MgSO4.7H2O

Again, the total number of electrons in one formula unit of MgSO4.7H2O
=1×12+1×16+14+11×8=130

So, the total number of electrons in one mole of MgSO4.7H2O
=130×6×1023

Therefore, the total number of electrons in 5.511 moles of MgSO4.7H2O
=130×6×1023×12=3.9×1025

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