Question

calculate the number of Aluminium ions present in 0.051 gram of Aluminium oxide​

Answer 1

Answer:

Step-by-step explanation:

Molecular weight of Al2O3 = 102 g

102 g of Al2O3 contains 6.022X1023 molecules

0.051g of Al2O3 contain X number of molecules

X = 0.051 X 6.022X1023 /102 = 0.003 X 1020

X = 3 X 1020

1 molecule of Al2O3 contains – 2 Al+3ions

3 X 1020 molecules contains – 2 X 3 X 1020

= 6 X 1020 Al+3 ions

∴0.051 g of Al2O3 contains 6 X 1020 Al+3 ions

Answer 2

Answer:- 0.051×10-²

Step-by-step explanation:

number of ions(n) = given mass of aluminium/molar mass of Al

   n = 0.051/100 = 0.00051 ions.

    or

n = 0.051×10-²ions