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Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) - Science

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

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

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Solution

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

i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains`=(6.022xx10^(23))/102xx0.051" molecules"`

                                                 = 3.011 × 1020 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 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020

= 6.022 × 1020

Concept: Concept of an Ion
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APPEARS IN

NCERT Class 9 Science
Chapter 3 Atoms and Molecules
Q 11 | Page 44
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