calculate the member of aluminium ions present in 0.051g of aluminium
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Calculate the number of aluminium ions present in 0.051g of aluminium oxide(Al2O3).
Molecular weight of Al2O3 = 102 g.102 g of Al2O3 contains 6.022X1023molecules.0.051g of Al2O3 contain X number of molecules.X = 0.051 X 6.022X1023 /102 = 0.003 X 1020X = 3 X 10201 molecule of Al2O3 contains – 2 Al+3ions.
Molecular weight of Al2O3 = 102 g.102 g of Al2O3 contains 6.022X1023molecules.0.051g of Al2O3 contain X number of molecules.X = 0.051 X 6.022X1023 /102 = 0.003 X 1020X = 3 X 10201 molecule of Al2O3 contains – 2 Al+3ions.
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102 grams is the answer
zoyasiddique:
explain please
already explanation is in another answer
where
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