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    17.1g alluminium sulfate al2(so4)3 dissolved in water calculate number of sulfate ions in so2- present in solution formed
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    assume molarmass al2(so4)3 is 342 gmol-1 and avogadro constant is 6x10 power 23gmol-1

    is it a 3x10power21
    b1x1022
    c3x10power22
    d9x10power22

    its d in mark scheme but how do u solve it please help asap its from exam pastpaper
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    It is correct that the answer is (d).
    You will calculate this by the following way:

    Moles of aluminum sulfate = 17.1/342 =0.05
    in the solution there are 3 sulfate ions.

    Hence number of sulfate ions
    0.05*3*6*10^23 = 9*10^22
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    What board?, and you should get this moved to the Chem forum
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    I concur with the above and sadly I can't help this question format is really different to AQA plus I'm terrible when it comes to Chemistry.
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    Al2(SO4)3 --> 2Al(3+) + 3SO4(2-) moles(Al2(SO4)3) = 17.1/342 = 0.05 moles. Ratio of Al2(SO4)3 to to SO4(2-) is 1:3 => moles of SO4(2-) = 0.05 x 3 = 0.15 mol. Avagadro constant = 6 x 10^23 mol-1. Therefore the number of sulphate ions are: 0.15mol x (6 x 10^23 mol-1) = 9 x 10^22 . Option D.
 
 
 
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