find the number of oxygen and Aluminium ions in 1.02 gram of Aluminium oxide
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Given;
W = 1.02 g
M = 54 + 48 = 102 g/mol.
let NA = 6 × 10²³ for approximate value
Thus moles of Al2O3 ;
- n = W/M
- n=1.02/102
- n=1/100 moles
1 moles = 2 Al3+ ions and 3 O2- ions
Thus moles of Al3+ ions;
- n×2 = 1/50mol.
Thus no. of Al3+ ions;
- 1/50 × NA = 1/50 × 6 × 10²³
- 0.12 × 10²³ ions
Thus moles of O2-;
- n×3 = 3/100 = 0.03 moles
Thus no of O2- ions;
- 0.03 × NA = 0.03 × 6 × 10²³
- 0.18 × 10²³ ions
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