Mole calculations revision
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Mole calculations revision. AQA Chem 1 Jan 2012. AQA Chem 1 Jan 2012. AQA Chem 1 Jun 2011. AQA Chem 1 Jun 2011. a) 5.2g of an acid HX were dissolved in 250cm 3 of solution (HX has an Mr of 204.1) Calculate the concentration of HX.

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Mole calculations revision

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Mole calculations revision


AQA Chem 1 Jan 2012


AQA Chem 1 Jan 2012


AQA Chem 1 Jun 2011


AQA Chem 1 Jun 2011


a) 5.2g of an acid HX were dissolved in 250cm3 of solution (HX has an Mr of 204.1)

Calculate the concentration of HX.

b) A 25.0 cm3 portion of HX was titrated against NaOH of unknown concentration; the accurate titres were 24.20cm3 23.95cm3 and 24.05cm3.

Calculate the concentration of the NaOH.


  • Moles HX = 5.2 / 204.1 = 0.02548

    Conc HX = 0.02548 x 1000 / 250 = 0.102 mol dm-3

    b) Moles HX = c x v / 1000 = 0.102 x 25 / 1000

    = 2.55 x 10-3

    HX + NaOH = NaX + H2O (i.e. 1:1 ratio)

    Moles NaOH = moles HX = 2.55 x 10-3

    VolNaOH = mean titre (two concordant titres)

    = 23.95 + 24.05 / 2 = 24.00

    ConcNaOH = n x 1000 / vol (mean titre)

    = 2.55 x 10-3 x 1000 /24 = 0.106 mol dm-3


OCR F321 May 2010


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