A solution contains 29.1g per litre of aluminium Sulphate. Calculate the number of Sulphate ions in 350cm^{3} of the solution.(Al = 27, S = 32, O = 16) Avogadro’s constant = 6.0 x 10^{23.}

Molar mass of Al_{2}(SO_{4})_{3} = 27 x 2 + (32 x 3) + (16 x 4 x 3) = 342Moles of Al_{2}(SO_{4})_{3} in 1000cm^{3} = Moles of Al2(SO_{4})_{3} in 350cm^{3} =Al_{2}(SO_{4})_{3(aq)} →2Al^{3+}_{(aq)} + 3SO_{4}^{2-}_{(aq)}1 mole of Al_{2}(SO_{4})_{3} produce 3 moles of SO_{4}^{2-} ions0.02979 moles → 0.02979 x 3 moles of SO42- ionsNumber of SO_{4}^{2-} ions = 0.08937 x 6.0 x 10^{23} = 5.36 x 10^{22} ions.

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