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Mole calculations revision

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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