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    I don't even know where to start off for both a) and b)

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    i got a) but dunno ow to do b)
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    On addition of iodide, the liberated copper(ii) cations react to form iodine (Eq. 1) which becomes the analyte of the titration.
    A mean titre of 20.50 cm3 is equivalent to 2.255e-3 mol of thiosulfate added.
    From the stoichiometry ratio (Eq. 2), the moles of iodine are half those of thiosulfate (1.1275e-3 mol).

    Back to Eq. 1, the amount of copper(ii) cations is twice the amount of iodine (2.255e-3 mol).

    Hydrated copper(ii) sulfate dissolves in water like this:
    CuSO4.nH2O(s) + aq → Cu2+(aq) + SO42-(aq) + nH2O(l)
    The amount of crystal is equal to the amount of copper(ii).
    However, since just 1/10 of the solution was used, the total amount of crystal is 2.255e-2 mol.

    2.255e-2 mol of CuSO4 has mass = 3.60 g. The rest out of 5.63 g must therefore be water (2.03 g = 0.11 mol).
    If 2.255e-2 mol carries 0.11 mol of water, n = 5. The crystal is pentahydrate.
 
 
 
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