Question

No. of electron present in
6.75gm of A^l3+ ion
(Atomic weight = 27)

​

Answer 1

Answer:

Number of electrons = 1.5055 × 10²⁴

Explanation:

Number of electron in Aluminium = 13

Now,

When we remove 3 electron from Al it becomes Al³⁺ ion.

Then the number of electrons in one Al³⁺ ion = 13 - 3 = 10   →→ (1)

Now,

Atomic mass of Al³+ ion = 27 g/mol.

[If we remove or add electron to any element, mass number doesn't change]

Given mass of Al³⁺ = number of 6.75 g

So,

Mole of Al³⁺ = 6.75/27 = 0.25 mol

Then,

Number of Al³⁺ ions in 0.25 mol = 0.25 × 6.022 × 10²³ = 1.5055 × 10²³

Now,

Number of electrons in 1.5055 × 10²³ Al³⁺ ions = 10 × 1.5055 × 10²³ [from (1)}

∴ Number of electrons = 1.5055 × 10²⁴