Calculate the number of aluminium ions present in 0.051 gram of aluminium oxide
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The chemical formula of Aluminium oxide is Al_2O_3.
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1 Mole of Al_2O_3=2.27+3.16=102g.
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Moles of Al_2O_3=0.051/102=5.10^{-4}
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5.10^{-4} g of Al_2O_3 contains= 5.10^{-4}.6,02.10^{23}=3,011.10^{20}
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Number of Al^{3+} ions in 1 molecule of Al2O3 is 2.
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The number of Aluminium ions in 0.051 g of Al_2O_3 will be,
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\large 2.(3,011).10^{20}=6,022.10^{20}
Answer: 6,022.10^{20}