Question

Calculate no of aluminium ions present in o.o51g of aluminium oxide

Answer 1

GIVEN MASS(m) = 0.051g
MOLE(n)=GIVEN MASS/MOLECULAR MASS
MOLECULAR MASS OF ALUMINUM =27
n= 0.051/27
ONE MOLES IS EQUAL TO
$6.022 \times {10}^{23}$
0.051/27 MOLES IS EQUAL TO
$(0.051 \div 27) \times 6.022 \times {10}^{23}$
The final answer is
$1.14 \times {10}^{21}$

Answer 2

☺ Hello mate__ ❤

◾◾here is your answer...

1 mole of Al2O3 = 2 × 27 + 3 × 16 = 102g

i.e.,

102g of Al2O3 = 6.022 × 10^23 molecules of Al2O3

Then, 0.051 g of Al2O3 contains

=(6.022×10^23/102) ×0.051 molecules of Al2O3

= 3.011 × 10^20 molecules of Al2O3

The no. Al^3+ in one molecules of Al2O3 is 2.

Here, Al^3+ = aluminium ion .

Therefore, The number of Al^3+ present in 3.11 × 10^20 molecules (0.051g) of Al2O3

= 2 × 3.011 × 10^20

= 6.022 × 10^20

I hope, this will help you.

Thank you______❤

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