Question

calculate the number of Aluminium ions present in 0.051 gram of Aluminium oxide​

Answer 1

$\bf\large\underline\pink{Solution :-}$

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

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)

❒_____________________

Answer 2

Answer:

(Al203) is the solution