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questin 8.5 new aqa specimen paper

Discussion in 'Science' started by urvipp1982, Apr 14, 2018.

  1. urvipp1982

    urvipp1982 New commenter

    Dear All, i am trying to help a neighbours daughter with her chemistry and for the life of me i cannot figure out where they got the 0.0027 from. converting cm3 into dm3 is /1000 which equates to 0.027. so am confused. Please help.

    The student carried out five titrations. Her results are shown in Table 5.

    A student used a pipette to add 25.0 cm3 of sodium hydroxide of unknown concentration to a conical flask.
    The student carried out a titration to find out the volume of 0.100 mol/dm3 sulfuric acid needed to neutralise the sodium hydroxide.

    Titration 1 27.40

    Titration 2 -28.15

    Titration 3 27.05

    Titration 4 27.15

    Titration 5 27.15

    Volume of 0.100 mol/dm3 sulfuric acid in cm3

    4 Concordant results are within 0.10 cm3 of each other.

    Use the student’s concordant results to work out the mean volume of 0.100 mol/dm3 sulfuric acid added.

    [2 marks] 27.12

    5 The equation for the reaction is:

    2NaOH + H2SO4 → Na2SO4 + 2H2O

    Calculate the concentration of the sodium hydroxide.
    Give your answer to three significant figures.

    [4 marks]


    mark scheme says :
    Moles H2SO4 = conc × vol =

    0.00271 - but how did they get this figure. am i being a complete nunz? as 27.12 /1000= 0.02712, where has the extra 0 come from?

    Ratio H2SO4:NaOH is 1:2 or

    Moles NaOH = Moles H2SO4 × 2

    = 0.00542

    Concentration NaOH = mol/vol

    = 0.00542/0.025 = 0.2168

    0.217 (mol/dm3)




  2. ah3069

    ah3069 Occasional commenter

    to calculate moles of sulphuric acid, you do the volume multiplied by the concentration

    so volume in cm3 is 27.12

    Volume in dm3 = 27.12 / 1000
    = 0.02712

    The to calculate moles, you multiply concentration and volume:

    concentration is 0.1 mol/dm3 from the information

    moles = 0.1 x 0.02712
    moles = 0.002712

    Then the rest of the working from there involves:

    molar ratio 1:2

    Moles of NaOH = 2 x 0.002712
    = 0.005424

    Concentration = moles / volume
    = 0.005424 / 0.025
    = 0.217 mol/dm3

    It may be easier to use the grid method of doing this, it seems to work better for the pupils I teach!

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