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    0.0859g of compound X which contains a group II metal dissolves in water to produce colourless solution.

    excess acidifed silver nitrate added. 0.308g of white precipitate (Y) was formed,

    What is Y? My answer = AgCl
    Compound X? My answer = CaCl2
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    I think it would have to be a lighter group 2 metal than Ca.

    0.308g of AgCl would be 0.00215 mol. Halving that because a grp2 compound would have two Cl, you need 0.00108 ish moles of your group 2 compound to produce this. Try the calculations with Mg and Be and see which fits.
 
 
 
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