Calculate the number of aluminium ions present in 0.051 g of aluminium

1.	Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Given: At. Mass of AL = 27 u)
Answer» Given: At. Mass of AL = 27 u Molar mass of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 54 + 48 = 102g This means 102g of Al₂O₃ has 2 × 6.022 × 10²³ molecules Then 0.051g contains: Number of Al³⁺ ions = \(\frac{Given\,mass}{Molar\,mass}\)x 2 x 6.022 x 10²³ Number of Al³⁺ ions = \(\frac{0.051g}{102g}\)x 6.022 x 10²³ ⇒ Number of Al³⁺ ions = 6.022 x 10²⁰ Thus.0.051 g of aluminium oxide contains 6.022 x 10²⁰ aluminium ions.

Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Given: At. Mass of AL = 27 u)

Answer»

Given: At. Mass of AL = 27 u

Molar mass of aluminium oxide (Al₂O₃) = 2 × 27 + 3 × 16 = 54 + 48 = 102g

This means 102g of Al₂O₃ has 2 × 6.022 × 10²³ molecules

Then 0.051g contains:

Number of Al³⁺ ions = \(\frac{Given\,mass}{Molar\,mass}\)x 2 x 6.022 x 10²³

Number of Al³⁺ ions = \(\frac{0.051g}{102g}\)x 6.022 x 10²³

⇒ Number of Al³⁺ ions = 6.022 x 10²⁰

Thus.0.051 g of aluminium oxide contains 6.022 x 10²⁰ aluminium ions.