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Check Concentration Calculation (Very Wordy As For Coursework)

  1. Mar 31, 2009 #1
    1. The problem statement, all variables and given/known data

    You are provided with a solution of hydrogen peroxide which is known to be approximately ‘100-volume’. Plan two experiments that would enable you to determine the exact concentration of H2O2 in mol dm-3, in this solution

    2. Relevant equations

    2H2O2 (aq) ----------------> O2 (g) + 2H2O (l)

    3. The attempt at a solution

    My friends solution

    The term ‘100-volume’ means that for every 1cm3 of H2O2, 100cm3 of oxygen gas will be liberated. As this volume of gas is not ideal for a chemical experiment, the H2O2 will have to be diluted by a factor of 10 to give is 10-volume H2O2 instead. This will be done by the following steps:

    1) Accurately measure 25cm3 of ‘100-volume’ H2O2 using a graduated pipette and empty into a 250cm3 conical flask.
    2) Fill the conical flask up to 250cm using distilled water

    The oxygen gas will be collected and measured by using gas syringe. As a typical gas syringe holds about 100cm3, a desirable amount of oxygen to collect would be about 60cm3. 10-volume H2O2 liberates 10cm3 of oxygen for every 1cm3 of H2O2 and so therefore 6cm3 of H2O2 will be required. From this the concentration of H2O2 will be calculated as follows:

    -1mole of gas = 24000cm3.
    -As 60cm3 of gas is collected, moles = 60/24000 = 0.0025moles
    -The stoichiometry of the equation states that for every mole of oxygen, there are twice as many of H2O2. Therefore moles of H2O2 = 2 x 0.0025 = 0.005moles
    -concentration = moles/volume (in dm3); 6cm3 = 0.006dm3.
    Concentration = 0.005/0.006 = 0.83moldm-3
    -As the original concentration of H2O2 was diluted by a factor of 10, the original concentration is 10times as much as 0.83moldm-3 and so therefore is 0.83 x 10 = 8.3moldm-3.

    My Solution

    ‘100-Volume’ means that 1cm3 of H2O2 will release 100cm3 of oxygen gas. Starting with a 100-Volume H2O2 would be unreasonable as a gas syringe can only hold up to 100cm3 of gas, therefore I am going to dilute the H2O2 by a factor of 10 making a ’10-Volume’ solution.
    This will be done by taking 25cm3 of 100-Volume H2O2 into a bulb pipette then putting it into a 250cm3 conical flask and then adding 225cm3 of distilled water thus creating a ’10-Volume’ solution. This now means 1cm3 of H2O2 will release 10cm3 of oxygen gas.

    Knowing a gas syringe can only hold up to 100cm3 of gas a good sample of gas to collect would be 70cm3 of oxygen. To get 70cm3 I would require 7cm3 of 10-Volume H2O2

    With this information I can calculate the concentration of 100-Volume H2O2

    Firstly I must convert 70cm3 to dm3 dividing through by 1000
    70/1000=0.07 dm3

    Secondly I must use the rule that one mole of any gas occupies 24dm3 at room temperature, this is to calculate the moles of O2.
    0.07/24=0.00292 moles

    I can now use this information to determine the moles of H2O2. Using the equation the moles of H2O2 to moles of O2 is 2:1 so
    Moles of H2O2 = 0.00292 x 2 = 0.00583 moles

    I can now calculate the concentration of 10-Volume H2O2.
    Concentration = (Moles/Volume) mol dm-3
    (0.00583)/0.07=0.083 mol dm-3

    This is the concentration of the 10-Volume solution to get the solution of the 100-Volume solution I must times 0.083 by 10 which gives a concentration 0.83 dm-3 of 100-Volume H2O2.

    So whos right me or my friend?
     
  2. jcsd
  3. Mar 31, 2009 #2

    Borek

    User Avatar

    Staff: Mentor

    What was the volume of the solution used?
     
  4. Apr 1, 2009 #3
    Ahh Thanks I was dividing by 0.07dm3 O2 not the 0.007dm3 of h2o2 T_T!
     
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