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.