Question

$\Huge\rm{{Question }}$
Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.
(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Class 9 Science .

Answer 1

Answer:

Mole of aluminium oxide (Al2O3) is

⇒ 2 x 27 + 3 x 16

Mole of aluminium oxide = 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.

Therefore, the number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3)

= 2 × 3.011 × 1020 = 6.022 × 1020

Answer 2

Answer:

6.023 × 10²⁰ ions

Step-by-step explanation:

Given :

0.051 g of Aluminium oxide

Atomic mass of Al = 27 u

To find :

the number of aluminium ions in the 0.051 g of aluminium oxide.

Solution :

We know that, the molecular formula of Aluminium oxide is Al₂O₃

The mass of 1 mole of Al₂O₃ is [Mass of Al = 27 ; O = 16]

=> 2(27) + 3(16)

=> 54 + 48

=> 102 g = mass of Al₂O₃

So, As the molecules of Aluminium in Al₂O₃ are 2,

they will contain 2 × 6.023 × 10²³ ions {For 1 mole}

Now, here, it is asked for 0.051 g of aluminium oxide.

=> 51 × 10⁻³ g

As there are 2 × 6.023 × 10²³ ions for 102g

So, how many for 51 × 10⁻³ g?

To know this, let us cross multiply,

Let us assume number of molecules as x for 51 × 10⁻³ g

$\implies \mathrm{\dfrac{2 \times 6.023 \times 10^{23}}{x} = \dfrac{102}{51 \times 10^{-3}}}$

$\implies \mathrm{\dfrac{2 \times 6.023 \times 10^{23}}{x} = \dfrac{\cancel{102}^2}{\cancel{51} \times 10^{-3}}}$

$\implies \mathrm{x = \dfrac{\not 2\times 6.023 \times 10^{23} \times 10^{-3}}{\not2}}$

$\therefore \underline{\boxed{\mathbf{x = 6.023 \times 10^{20}}}}$

Thus, 6.023 × 10²⁰ ions of aluminium are present in 0.051 g of aluminium oxide.

Hope it helps!!