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Calculate the number of aluminium ions present in $0.051\,g$ of aluminium oxide.
[HintThe mass of an ion is the same as that of an atom of the same element. Atomic mass of $Al=27\,\mu $.]
$6.022 \times 10^{22} $
$6.022 \times 10^{16} $
$6.022 \times 10^{23} $
$6.022 \times 10^{20} $
Solution
Molar mass of $Al _{2} O _{3}=(27 \times 2)+(16 \times 3)=54+48=102\,g$
$\underset{\begin{smallmatrix}
1\,mol \\
(102\,g)
\end{smallmatrix}}{\mathop{A{{l}_{2}}{{O}_{3}}}}\,\,\rightleftharpoons \,\underset{2\,mol}{\overset{3+}{\mathop{2Al}}}\,\,+\,\overset{2-}{\mathop{3O}}\,$
$\because $ $102\, g $ $Al_{2} O _{3}$ contains $Al^{3+}$ ions $=2 \times 6.022 \times 10^{23}$
$\therefore $ $ 0.051 \,g$ $Al_{2} O _{3}$ will contain $Al^{3+}$ ions $=$$\frac{2\,\times \,6.022\,\times {{10}^{23}}}{102}\times 0.05$
$=6.022 \times 10^{20} \,Al^{3+}$ ions
