Question

calculate the number of Aluminium ions present in 0.051 gram of Aluminium oxide

Answer 1

answer is

Molecular weight of Al2O3 = 102 g

102 g of Al2O3 contains 6.022X1023 molecules.

Answer 2

Answer:

6.022*10^20

Explanation:

1 mole of Al2O3=formula mass of al2o3 in grams

= mass of al*2+mass of o*3

=27*2+16*3

= 54+48

=102 grams

Now, 1 mole of al2o3 contains 2 moles of Al.

so, mass of al in 1 mole of al2o3=mass of al x2

                                                    = 27*2

                                                     =54 grams

now, 102 g aluminium oxide contains= 54 g al

so, 0.051 g aluminium oxide contains= 54/102 x 0.051 g al

                                                             = 0.027 g al

now, 27 g of aluminium has ions= 6.022*10^23

so, 0.027g of aluminium has ions=6.022*10^23/27*0.027

                                                      = 6.022* 10^20

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