Find the no. of aluminium ions present in 0.051g of aluminium oxide??

1.	Find the no. of aluminium ions present in 0.051g of aluminium oxide??
Answer» 1 mole of aluminium oxide (Al2O3) = 227 + 3 16= 102 gi.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3Then, 0.051 g of Al2O3 contains == 3.011 * 1020 molecules of Al2O3The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.Therefore, the number of aluminium ions (Al3+) present in 3.011 * 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 * 3.011 * 1020= 6.022 * 1020

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