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    did i get this right and if not can u show how to get the full marks2 A solution of ethanedioic acid was prepared by dissolving 1.01 g of the acid in water and making the volume up to 250 cm3 in a volumetric flask. A 25.0 cm3 sample of this solution was titrated against NaOH and required 18.40 cm3 NaOH for neutralisation.
    (COOH)2  2NaOH → (COONa)2  2H2O
    a Calculate the number of moles of ethanedioic acid in 1.01 g. (1 mark)
    b Calculate the concentration of the ethanedioic acid solution. (1 mark)
    c Calculate the number of moles of ethanedioic acid used in the titration and hence calculate the number of moles of NaOH required for neutralisation.
    d Calculate the concentration of the NaOH solution in:
    i mol dm−3
    ii g dm−3
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    this is my answer and workin out
    mas/mr =1.01/90=0.0112
    conc=massXvolume =0.0112/(250/1000)=0.448x2 = 0.896
    0.896/0.025=35.84
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    hi, did you find the answer to this?
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    (Original post by loregillis99)
    hi, did you find the answer to this?
    nah cant remember prob not
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    (Original post by jkjj)
    this is my answer and workin out
    mas/mr =1.01/90=0.0112
    conc=massXvolume =0.0112/(250/1000)=0.448x2 = 0.896
    0.896/0.025=35.84
    You've only given 3 answers to 4Q's. If you can order it properly I might be able to help u.
 
 
 
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