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

#### Answers

** Given mass = 0.051 g. **

** Let the given mass be 'm ' **

** Formula of compound 'Aluminium Oxide ' = Al_{2} O_{3 .} **

** 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 'n_{0} **

**' = 6.022 * 10**^{23}. * Then n = ( m* n_{0}) / M *

* = ( 0.051 * 6.022 * 10^{23 }) / 102 *

* = *

*0.307122 * 10*^{23}/ 102 ** = 0.003011 * 10^{23 }ions. **

** = **

*3.011 * 10*^{21 }ions.the above one is wrong because he has found out the no. of ions in Al_{2}O_{3}but 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 * 10^{20}

but you multiply it by 2

so ans will be 6.022 * 10^{20}

*1 mole of aluminium oxide (Al _{2}O_{3}) = 2 × 27 + 3 × 16*

*= 102 g*

*i.e., 102 g 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 = *

*= 3.011 × 10 ^{20} molecules of Al_{2}O_{3} *

*The number of aluminium ions (Al ^{3+}) present in one molecule of aluminium oxide is 2.*

*Therefore, the number of aluminium ions (Al ^{3+}) present in 3.011 × 10^{20} molecules (0.051 g ) of aluminium oxide (Al_{2}O_{3}) = 2 × 3.011 × 10^{20} *

*= 6.022 × 10 ^{20} *

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** 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... **

Thanks akshaygahlot and shobhitanshu.

But how did that 2 come from? (u multiplied 3.011 x 10^{20} by two to get the last answer)