Question

calculate the no of aluminum ions present in 0.051gof aluminium oxide​

Answer 1

Answer:

6.022 × $10^{20}$

Explanation:

1 mole of Aluminium Oxide () = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of Aluminium Oxide ($Al_{2}$$O_{3}$) = 6.022 × $10^{23}$ molecules of Aluminium Oxide (

We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number

Then, 0.051 g of Aluminium Oxide () contains

= (0.051 ÷ $10^{20}$ ) × 6.022 × $10^{23\\}$ molecules of Aluminium Oxide ($Al_{2}$$O_{3}$)  

= 3.011 × molecules of Aluminium Oxide ()

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

Therefore, the number of Aluminium ions () present in 3.011 × molecules (0.051 g ) of Aluminium Oxide ()

= 2 × 3.011 ×  $10^{20}$

= 6.022 x $10^{20}$

Answer 2

Answer:

your answer

Explanation:

1 mole of Aluminium Oxide (Al2O3) = 2 × 27 + 3 × 16  = 102 g

We know, 102 g of Aluminium Oxide (Al2O3) = 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)

We know, The number of atoms present in given mass = (Given mass ÷ molar mass) × Avogadro number

Then, 0.051 g of Aluminium Oxide (Al2O3) contains

= (0.051 ÷ 102 ) × 6.022 × 1023 molecules of Aluminium Oxide (Al2O3)  

= 3.011 × 1020 molecules of Aluminium Oxide (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

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