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)

Molar mass of aluminium oxide Al203
= (2 x 27) + (3 x 16)
= 54 + 48 = 102 g.
Therefore, 102g of Al203 contains = 2 x 6.022 x 1023 aluminium ion
 Therefore,0.051 of Al203 contains = (2 x 6.022x 1023)/102 x 0.051
                                                                = (12.044 x 1023 x 0.051)/102
                                                                = (0.614 x 1023)/102
                                                                = 0.006022 x 1023
                                                                = 6.022 x 1020 Al+3 ions.

[Extra information: Molar mass is defined as the mass of a substance for a given amount where the amount of the substance refers to the number of molecules or atoms or compounds present in one mole of a substance. The SI unit of the Molar mass of a substance is always calculated in g/mol.
The chemical formula of the compound is a symbolic representation of its composition, e.g., the chemical formula of sodium chloride is NaCl.
It indicates the elements that are present in the compound and the ratio of these elements in the compound.]
 

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Updated on: 14-Mar-2023

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