Question

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
Please answer properly as per NCERT.

Answer 1

Answer:

Mole of aluminium oxide (Al2O3) = 2 x 27 + 3 x 16

= 102 g

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

Then, 0.051 g of Al2O3 contains = 6.022 x 1023 / 102 x 0.051 molecules

= 3.011 x 1020 molecules of Al2O3

The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Answer 2

Answer:

molar mass of Al2O3= 27×2 + 16×3

=102g

no.of mole = given mass/molar mass

= 0.051/102

= 0.0005

no.of mole = no.of ions/avagrado no.

0.0005=N/No

N=0.0005×6.022×10^23

=