Question

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Atomic mass of aluminium = 27u)

Answer 1

Hint: Attempt this question by writing the formula of aluminium oxide, find its molar mass. Then use Avogadro’s number to find the answer.

Complete step by step answer:
The formula of Aluminium oxide is \[A{{l}_{2}}{{O}_{3}}\], i.e. it contains 2 atoms of aluminium and 3 atoms of oxygen.
So, Molar mass of Aluminium oxide = 2 x mass of aluminium + 3 x mass of oxygen
 = 2 x 27 + 3 x 16
= 102 g
Avogadro’s number is the number of particles (electrons or molecules or atoms) in 1 mole (or mol) of a substance. The value of Avogadro’s number is approximately \[6.022\text{x}{{10}^{23}}\text{mo}{{\text{l}}^{-1}}\].
Therefore, 102 g of aluminium will contain = \[=\dfrac{6.022\text{x}{{10}^{23}}}{102}\]\[A{{l}^{3+}}\]ions
0.051 g of aluminium oxide will contain \[=\dfrac{6.022\text{x}{{10}^{23}}}{102}\text{x}0.051\]
\[=6.022\text{x}{{10}^{20}}A{{l}^{3+}}\text{ions}\text{.}\]
Therefore, the answer is:
The number of aluminium ions (\[A{{l}^{3+}}\]) present in one molecule of aluminium oxide, \[A{{l}_{2}}{{O}_{3}}\] is \[=6.022\text{x}{{10}^{20}}A{{l}^{3+}}\text{ions}\text{.}\]

Additional information: Atomic mass unit is defined as “the 1/12th weight of the mass of one carbon atom”.
Therefore, \[1\text{ amu }=\text{ }1.66\text{x}{{10}^{-24\text{ }}}\text{grams}.\]
Also, the mass of an ion of an element is the same as that of an atom of the same element. Therefore, the mass of any ion of aluminium will be 27u, regardless of the valency.

Note: Aluminium oxide is also known as alumina. The most commonly (naturally) occurring crystalline form of aluminium oxide is ‘Corundum’. It is a thermodynamically stable crystal. The structure of crystalline aluminium oxide is octahedral. It has many uses – as a filler in plastics, ingredient in glass, catalyst for Claus process, for gas purification, etc.