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# as chemistry unit 1 heeeeeeeelp please Watch

1. 17.1g alluminium sulfate al2(so4)3 dissolved in water calculate number of sulfate ions in so2- present in solution formed
4

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
2. 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
3. What board?, and you should get this moved to the Chem forum
4. 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.
5. 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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