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Calculate the no. of aluminium ions present in 0.051 gm of Aluminium oxide.

Asked by Rajat Narsapurk...(student) , on 9/12/13


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EXPERT ANSWER

Aluminum oxide has the molecular formula of Al 2 O 3 .

Atomic mass of aluminum = 26.98 g/mol

Atomic mass of oxygen = 16 g/mol

The molecular mass of aluminium oxide is = (2 × 26.98) + (3 ×16)

                                                                 = 101.9648 g/mol

Number  of moles = given mass/ molar mass

Number of moles in 0.051g of Al 2 O 3 = 0.051/101.9648

                                                         = 0.0005 mol

As 1 mole  Al 2 O 3 contains 2 mole Al.

So, 0.0005 mole of Al 2 O 3 contains = 0.0005×2

                                                       = 0.001mol Al

As 1 mole = 6.022 ×10 23 ions

Therefore, 0.001mol of Al 2 O 3 = 6.022 × 10 23 × 0.001

= 6.022 × 10 20 ions

Hence 0.051 g of Al 2 O 3 contains 6.022×10 20 ions.

Posted by Priya Singh(MeritNation Expert)on 12/12/13

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