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    2014 Question 1 (f) Aimeeohara

    Can someone please tell me where the solutions are getting 1.14g of anhydrous sodium carbonate? I can't figure it out!

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      Weam

      You can check tha marking scheme ...

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      Aimeeohara

      It just says 1.14g on the marking scheme. I dont know where they're getting it from.

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      BarryCheung

      (ii) 0.0108 x 106g (Mr of Na2CO3) = 1.1448g/L

      (f)

      Mass of water of crystallisation

      sample of crystals were 2.50g and you have found the grams per litre above...

      = 2.50 - 1.1448 = 1.3552g

      % water of crystallisation = 1.3552/2.50 x 100 = 54.21%

      x = moles of H2O/moles of Na2CO3

      = 0.0752/0.0108 = 6.96

      x = 7

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      Me

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