Question

Q.1. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
[HintThe mass of an ion is the same as that of an atom of the same element. Atomic mass of AI= 27 u.].






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Answer 1

Explanation:

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 102 g

i.e., 102 g of Al2O3= 6.022 x 10^23 molecules of Al2O3

Then, 0.051 g of Al2O3 contains = 6.022 x 10^23 / (102 x 0.051 molecules)

= 3.011 x 10^20 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 10^20 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 10^20 = 6.022 × 10^20