Q. 11. Calculate the number of aluminium ions present in 0.051 g of aluminium
oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic
mass of Al = 27 u.]
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Answer:
1 mole of Al
O
2
3
=2×27+3×16
=102g
Moles of Al
2
O
3
=
102
0.051
=5×10
−4
moles
1 mole contains 6.02×10
23
molecules
∴5×10
−4
g of Al
2
O
3
contains = 5×10
−4
×6.02×10
23
=3.011×10
20
molecule
no. of Al
3+
ions in 1 molecules of Al
2
O
3
is 2.
∴ No. of aluminium ions will be:-
2×3.011×10
20
=6.022×10
20
.
Explanation:
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