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

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

.

h let me think

explanation:

i think option no d is the correct answer