Calculate the number of Aluminium ions present in 0.051g of Aluminium oxide
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains =
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
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the above one is wrong because he has found out the no. of ions in Al2O3but didn't find no. of ions in aliminium
to find that you just multiply the ans by 2
and he has done calculation wrong according to him ans must be
3.011 * 1020
but you multiply it by 2
so ans will be 6.022 * 1020
Thanks akshaygahlot and shobhitanshu.
But how did that 2 come from? (u multiplied 3.011 x 1020 by two to get the last answer)
I am sorry I typed incorrectly as i was typing without seeing the keyboard and by mistake i clicked on Post Answer button after solving only half question. Your answer is right akshaygahlot13...
Given mass = 0.051 g.
Let the given mass be 'm'
Formula of compound 'Aluminium Oxide' = Al2 O3 .
Formula mass = 102 u ( Al = 27 u and O = 16 u.)
Let the Formula mass be 'M'
Let the number of Aluminium ions present be 'n'.
Let Avagadro no. be 'n0 ' = 6.022 * 1023.
Then n = ( m* n0) / M
= ( 0.051 * 6.022 * 1023 ) / 102
= 0.307122 * 1023 / 102
= 0.003011 * 1023 ions.
= 3.011 * 1021 ions.