Question

how many atoms are contained in 2.70g of aluminum provided that 32g of sulphur equals 6.02 × 10^(23)atoms​

Answer 1

`1.63 * {10}^(24)`

one atom of an element = 6.02 \times {10}^{23} atom

The mass of one atom of sulphur = 32g

The mass of one atom of aluminium = 27g

so one atom of aluminium = 6.02 \times {10}^{23}

27g of AL = 6.02 \times {10}^{23} atom

2.70g of AL = X atoms

Then you cross multiply ........

and get the answer

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