Question

The number of aluminium ions present in 0.251 g of al2o3 are

Answer 1

Hello Dear.

Given ⇒ 
Mass of the Aluminium = 0.251 g.

Molar Mass of the Al₂O₃(Aluminium Oxide) = 102 g/mol.
[∵ 27 × 2 + 16 × 3 = 54 + 48] = 102]

Now, Using the Formula,

No. of Molecules = (Mass/Molar Mass) × Avogadro's Number.
= (0.251/102) × 6.023 × 10²³
= 0.0024 × 6.023 × 10²³
= 0.0148 × 10²³ Molecules.


∵ 1 molecules of the Al₂O₃ contains 2  Al³⁺ ions.
∴ 0.0148 × 10²³ molecules of Al₂O₃ contains 2 × 0.0148 × 10²³ Al³⁺ ions.
   = 0.0296 × 10²³ ions.
  


∴ Number of ions of the Al³⁺ in the 0.251 g of the Al₂O₃  is 0.0296 × 10²³ ions.


Hope it helps

Answer 2

Answer: 2.96* 10^21

Explanation:

Given moleculer mass of Al2O3 is 102

since, 102 gm of Al2O3 =1 mole

1 gm of Al2O3=1/102

0.251gm of Al2O3 =1/102*0.251 mole

therefore, no. of molecules of Al2O3=0.251/102*6.022*10^23 molecule

No. of Aluminium ions present= 2*0.251/102*6.022*10^23=2.96*10^21