Question

Calculate the no. of aluminium and chloride ions present in 0.051 g of aluminium chloride

Answer 1

1 mole of aluminium oxide (Al2O3) = 2*27 + 3 * 16

= 102 g

i.e., 102 g of Al2O3 = 6.022 * 1023 molecules of Al2O3

Then, 0.051 g of Al2O3 contains =

= 3.011 * 1020 molecules of Al2O3

The 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

Plzzz mark it as brainliest

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______❤

✿┅═══❁✿ Be Brainly✿❁═══┅✿