Question

. Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
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Answer 1

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1 mole of aluminium oxide (Al2O3) = 2×27+3×16= 102 g.

We know, 102 g. of aluminium oxide (Al2O3) = 6.022×10²³ molecules of aluminium oxide (Al2O3)

We know, the number of atoms present in the given mass = (given mass ÷ molar mass )× Avogadro number

Then, 0.051 g. of aluminium oxide (Al2O3) contains

= (0.051÷102)×6.022×10²³ molecules of aluminium oxide (Al2O3)

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

therefore, the number of ions (Al3+) present in 3.011 ×10²³ molecules (0.051g) aluminium aluminium oxide (Al2O3)

= 2×3.011×10²³

= 6.022×10²³

Answer 2

Answer:

Explanation:

hope it helps