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calculate the number of aluminium ion present in 0.051g of aluminium oxide.
question from class 9 NCERT ch 3 back exercise. plz give a full explanation. plz overlook the question if you don't know the answer.
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the number of aluminium ion present in 0.051g of aluminium oxide =
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geetika3:
in the second step of answer why the Avogadro no. is multiplied by 2??
No. of Al(3+) ions in one molecule of Al2O3 = 2
so thats why its multiplied by 2
yup
ok
thank you so much
u r wlcm
why it is multiplied by 2
Answered by
2
molecular formula aluminum oxide =Al2O3
molecular mass of Al2O3 = 102 g
no of mole of Al2O3 = given wt/molecular wt
= 0.051/102 = 1/2 x 10^-3
= 5 x 10^-4
in one mole of Al2O3 , 2mole of Al+3 ions present .
so , 5 x 10^-4 mole of Al2O3 , 2 x 5 x 10^-4 = 10^-3 mole of Al+3 ions present .
we know ,
no of ions = no of mole × 6.023 x 10^23
so,
no of ions of Al+3 in Al2O3 = 10^-3 x 6.023 x 10^23 = 6.023 x 10^20
molecular mass of Al2O3 = 102 g
no of mole of Al2O3 = given wt/molecular wt
= 0.051/102 = 1/2 x 10^-3
= 5 x 10^-4
in one mole of Al2O3 , 2mole of Al+3 ions present .
so , 5 x 10^-4 mole of Al2O3 , 2 x 5 x 10^-4 = 10^-3 mole of Al+3 ions present .
we know ,
no of ions = no of mole × 6.023 x 10^23
so,
no of ions of Al+3 in Al2O3 = 10^-3 x 6.023 x 10^23 = 6.023 x 10^20
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