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    6) A student carries out an experiment to identify an unknown carbonate.
    • The student weighs a sample of the solid carbonate in a weighing bottle.
    • The student tips the carbonate into a beaker and weighs the empty weighing bottle.
    • The student prepares a 250.0 cm3 solution of the carbonate.
    • The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.

    (a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and thesolution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.

    The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acidto a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.

    The equation below represents the reaction between the carbonate and hydrochloric acid.

    M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)

    (i) Calculate the amount, in mol, of M2CO3 used in the titration.

    ii) The student’s mass readings are recorded below.
    Mass of weighing bottle + carbonate / g 14.92
    Mass of weighing bottle / g 13.34

    Use the student’s results to identify the carbonate, M2CO3.

    The answer for (i) was 0.00111. How? I can do (ii) obviously if I knew the answer to (i) because that's just rearranging n=M/mr.

    So can someone please explain how to get the answer? Thanks.

    http://www.ocr.org.uk/Images/171752-...-materials.pdf

    That's the paper and the mark scheme is on the bottom - question 6.
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    (Original post by _Lizabeth_)
    6) A student carries out an experiment to identify an unknown carbonate.
    • The student weighs a sample of the solid carbonate in a weighing bottle.
    • The student tips the carbonate into a beaker and weighs the empty weighing bottle.
    • The student prepares a 250.0 cm3 solution of the carbonate.
    • The student carries out a titration using 25.0 cm3 of this solution measured using a pipette with0.100 mol dm−3 hydrochloric acid in the burette.

    (a) The sample of carbonate is dissolved in approximately 100 cm3 of distilled water in a beaker and the solution transferred to a volumetric flask. The volume of the solution is made up to 250.0 cm3 withdistilled water.

    The student carries out the final part of the experiment by adding 0.100 mol dm−3 hydrochloric acid to a burette and performing a titration using a 25.0 cm3 sample of the aqueous carbonate. The titre is 22.20.

    The equation below represents the reaction between the carbonate and hydrochloric acid.

    M2CO3(aq) + 2HCl(aq) → 2MCl(aq) + CO2(g) + H2O(l)

    (i) Calculate the amount, in mol, of M2CO3 used in the titration.

    ii) The student’s mass readings are recorded below.
    Mass of weighing bottle + carbonate / g 14.92
    Mass of weighing bottle / g 13.34

    Use the student’s results to identify the carbonate, M2CO3.

    The answer for (i) was 0.00111. How?
    moles = molarity x volume (for the HCl)

    Then use the equation and you see that the moles of carbonate = moles of HCl/2

    I can do (ii) obviously if I knew the answer to (i) because that's just rearranging n=M/mr.

    So can someone please explain how to get the answer? Thanks.

    http://www.ocr.org.uk/Images/171752-...-materials.pdf

    That's the paper and the mark scheme is on the bottom - question 6.
 
 
 
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