Question

calculate the total number of ions present in 0.51g of aluminium oxide.

Answer 1

Here we go......

$Molar \: Mass \: of \: Alluminium \: Oxide \\ \\ = 2 \times 27 + 3 \times 16 \\ \\ = 54 + 48 \\ \\ = 102g \\ \\ 102g \: of \: Alluminum \: Oxide \: contains \: \\ \\ = 2 \times 6.022 \times {10}^{23} \: {Al}^{3 + } Ions \\ \\ Therefore \: 0.51g \: of \: Alluminium \: Oxide \: contains \: \\ \\ = \frac{2 \times 6.022 \times {10}^{23} }{102} \times 0.51g \\ \\ = 6.022 \times {10}^{20} \: {Al}^{3 + } \: Ions \: \\ \\ Hit \: \: a \: \: thnx \: if \: u \: gain \: help.$

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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