if a sample of cuso4.5h2o contains 96g of oxygen, then the moles of H-atoms present in the sample is:
(a) 12 (b) 15 (c) 10/3 (d) 20/3
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1 mole of CuSO₄.5H₂O has:
(4 + 5) × 16 = 144 g
Then, 96 g of oxygen is present in: 96/144 = 2/3 moles
now, 1 mole of CuSO₄.5H₂O has 5 × 2 = 10 moles of hydrogen atoms
So, there will be: 10 × 2/3 = 20/3 moles of H- atoms are present
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