Question

Calculate the number of aluminium ions (Al3+) in 0.056 g of alumina (Al2O3).

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Answer

Molecular mass of alumina (Al203) = 2 × Al3+ + 3 × 02-

= 2 × 27 u + 3 × 16u

= 102 u

Gram molecular mass = 102 g

1 mol of alumina (Al203) = 102 g

102 g of Al2O3 = 1 mol

0.056 g of Al2O3 = ( 1 × 0.056/102) mol

= 5.49 x 10-4 mol

We know that one mol of alumina contains 2 mol of Al3+ ions.

5.49 x 10-4 mol of Al2O3 contains 2 x 5.49 x 10-4 mol of Al3+ ions

Number of Al3+ ions in 0.056 g = 2 × 5.49 × 10-4 x 6.022 x 1023

                                                          = 6.613 x 1020 ions of AI3+

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