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Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u) |
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Answer» Correct Answer - `6.022 xx 10^(20)` ions mole of aluminium oxide `(Al_(2)O_(3))` = 2 × 27 + 3 × 16 = 102g i.e., 102g of `Al_(2)O_(3) = 6.022 × 10^(23)` molecules of `Al_(2)O_(3)` Then, 0.051 g of `Al_(2)O_(3)` contains `=6.022xx10^(23)//102xx0.051` molecules `= 3.011 × 10^(20)` molecules of `Al_(2)O_(3)` The number of aluminium ions `(Al^(3+))` present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions `(Al^(3+))` present in `3.11 × 10^(20)` molecules (0.051g) of aluminium oxide `(Al_(2)O_(3)) = 2 × 3.011 × 10^(20)` `= 6.022 × 10^(20)` |
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