Given that 32.0g sulphur contains 6.02 x 10 23 sulphur atoms how many atoms are there in 2.70g of aluminium? {Al = 27, S =32}
- 6.02 x 10 23
- 3.01 x 1022
-
6.02 x 1022
- 5. 08 x 1022
- 3.01 x 1022
The correct answer is: C
Explanation
27g Aluminium = \(6.02 \times 10^{23}\)
1g Al = \(\frac{6.02 \times 10^{23}}{27}\)
2.7g Al = \(\frac{6.02 \times 10^{23}}{27} \times 2.7\)
= \(6.02 \times 10^{22} Al \) atoms