Question

Calculate the number of aluminum ions present in 0.051 g of aluminum oxide.

Answer 1

Answer:

The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) . The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.

Hence you can calculate.

By

Aaditya Singh

Answer 2

Solution :

First of all, we'll find out the mass of aluminum atoms in 0.051 g of aluminum oxide

★ 1 mole of Al₂O₃ = Formula mass of Al₂O₃ in grams

→ Mass of Al × 2 + Mass of O × 3

→ 27 × 2 + 16 × 3

→ 54 + 48

→ 102 grams

Now, 1 mole of Al₂O₃ contains 2 moles of Al

So, mass of Al in 1 mole of Al₂O₃ = Mass of Al × 2

→ 27 × 2

→ 54 grams

Now, 102 g aluminum oxide contains 54 g Al

So,

0.051 g aluminum oxide contains $\sf\dfrac{54}{102}\times 0.051 g$ Al

→ 0.027 g

The atomic mass of aluminum is 27 u. 1 mole of aluminum atoms has a mass of 27 g, and it contains 6.022 × 10²³ aluminum ions.

Now, 27 g of aluminum has ions = 6.022 × 10²³

So,

0.027 g of aluminum has ions = 6.022 × 10²³/27 × 0.027

→ 6.022 × 10²⁰

$\therefore$ The number of aluminum ions in 0.051 grams of aluminum oxide is 6.022 × 10²⁰.