Question
How many number of aluminium ions are present in $$0.051\,g$$ of aluminium oxide ?
A.
$$6.023 \times {10^{20}}\,ions$$
B.
$$3\,ions$$
C.
$$6.023 \times {10^{23}}\,ions$$
D.
$$9\,ions$$
Answer :
$$6.023 \times {10^{20}}\,ions$$
Solution :
$${\text{Molar mass of}}\,\,A{l_2}{O_3}$$ $$ = 2 \times 27 + 3 \times 16 = 102\,g$$
$$0.051\,g\,\,{\text{of}}\,\,A{l_2}{O_3}$$ $$ = \frac{{0.051}}{{102}} = 0.0005\,mol$$
$$1\,mol\,\,{\text{of}}\,\,A{l_2}{O_3}\,\,{\text{contains}}$$ $$2 \times 6.023 \times {10^{23}}\,\,A{l^{3 + }}\,ions$$
$$0.0005\,mol\,\,{\text{of}}\,\,A{l_2}{O_3}\,\,{\text{contains}}$$ $$2 \times 0.0005 \times 6.023 \times {10^{23}}$$
$$A{l^{3 + }}\,ions = 6.023 \times {10^{20}}\,A{l^{3 + }}\,ions$$