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Keerthana Vijayan from CHINMAYA VIDYALAYA, asked a question
Subject: Science , asked on 28/1/12

Calculate the number of Aluminium ions present in 0.051g of Aluminium oxide

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Kunal Choudhary From St Cecilia's Public School, added an answer, on 30/1/12
33 helpful votes in Science

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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Akshay Gahlot From St Pauls Sr Sec Schoolbeawar, added an answer, on 28/1/12
41 helpful votes in Science

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

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Shobhitanshu From D A V Public School, added an answer, on 28/1/12
23 helpful votes in Science

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.

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Keerthana Vijayan From Chinmaya Vidyalaya, added an answer, on 29/1/12
135 helpful votes in Science

Thanks akshaygahlot and shobhitanshu.

But how did that 2 come from? (u multiplied 3.011 x 1020 by two to get the last answer)

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Shobhitanshu From D A V Public School, added an answer, on 29/1/12
23 helpful votes in Science

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...

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