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    How do you get to these answers?

    1. Ammonia is bought as "880 ammonia" (a solution of density 0.880g cm-3), whose content is 28% ammonia (i.e. it contains 280g ammonia per kilogram of solution). What volume of the "880 ammonia" solution would you need to prepare 1.0dm3 of 2.0 mol dm-3 ammonia solution?

    Answer was 0.138

    2. Caluclate the concentration (in mol dm-3) of 3% (percentage by mass) solution of potassium iodide. The density of the solution is assumed to be 1. Relative atomic masses: K=39, I=27.

    Answer was 0.18mol dm-3

    3. For the reaction, H2 + 3H2 <=> 2NH3, calculate the equilibrium constant if the concentrations at equilibrium are 0.082mol/1 N2, 2.14mol/1 H2, 0.26mol/1 NH3.

    Answer was 0.18
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    (Original post by Chqr)
    How do you get to these answers?

    1. Ammonia is bought as "880 ammonia" (a solution of density 0.880g cm-3), whose content is 28% ammonia (i.e. it contains 280g ammonia per kilogram of solution). What volume of the "880 ammonia" solution would you need to prepare 1.0dm3 of 2.0 mol dm-3 ammonia solution?

    Answer was 0.138

    2. Caluclate the concentration (in mol dm-3) of 3% (percentage by mass) solution of potassium iodide. The density of the solution is assumed to be 1. Relative atomic masses: K=39, I=27.

    Answer was 0.18mol dm-3

    3. For the reaction, H2 + 3H2 <=> 2NH3, calculate the equilibrium constant if the concentrations at equilibrium are 0.082mol/1 N2, 2.14mol/1 H2, 0.26mol/1 NH3.

    Answer was 0.18
    Question 1:
    From
    Moles = Vol x conc

    Moles of NH3 required is 2
    Mass required =17.031 x 2 = 34.062

    From data given you know 1kg of solution gives you 280g of NH3
    or 1g of NH3 per 3.57g of solution.

    Therefore you need 34.062g of NH3 which = (34.062 x 3.57) = 121.6g of solvent however since the density of the solvent is 0.88g cm^-3

    You need to allow for that deficit by multiplying g of solvent needed by 1.12 (1.12 x121.6 = 136cm^3 or 0.136 dm^3 of solution


    Question 2:

    Assuming you have 1dm^3 of solvent as it's 3% by mass this means there are 30g of KI per Kg. Thus moles=30/(127+39) = 0.18 and since it's in 1dm^3 and density is 1: there is 0.18mol dm^-3

    Question 3:
    Could you perhaps rewrite this question there seems to be a few mistakes and the answer I get is a bit away from the answer you state.
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    (Original post by haydyb123)
    Question 3:
    Could you perhaps rewrite this question there seems to be a few mistakes and the answer I get is a bit away from the answer you state.
    What answer did you get. The mark scheme could have a typo as its a word document from a 3rd party website and the answer is the same for Q2 and 3.

    So, yeah, what was your answer for Q3?
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    (Original post by Chqr)
    What answer did you get. The mark scheme could have a typo as its a word document from a 3rd party website and the answer is the same for Q2 and 3.

    So, yeah, what was your answer for Q3?

    Okay what I did;

    Kc= [NH3]^2/[N2][H2]^3

    And that gave me something like 0.84 but the units don't correlate either I make it to be 0.84 dm^6 mol^-2
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    I still dont get it how I am supposed to calculate those. I am sitting the test tomorrow and really struggling with those questions!!!
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    (Original post by Guddi)
    I still dont get it how I am supposed to calculate those. I am sitting the test tomorrow and really struggling with those questions!!!
    I've just worked out the first one differently to the other guy; the problem with these questions is that they're so stupidly worded.

    (Original post by Chqr)
    How do you get to these answers?

    1. Ammonia is bought as "880 ammonia" (a solution of density 0.880g cm-3), whose content is 28% ammonia (i.e. it contains 280g ammonia per kilogram of solution). What volume of the "880 ammonia" solution would you need to prepare 1.0dm3 of 2.0 mol dm-3 ammonia solution?

    Answer was 0.138
    When it talks about preparing 1.0dm3 of 2.0 mol dm-3 ammonia solution, it actually means what volume of 880 ammonia solution do you need to have the same amount of ammonia as in 1.0dm3 of a 2.0 mol dm-3 solution.

    1) We are told the second solution is 2 moles per dm3, as we have 1.0dm3, 2 moles of ammonia is present
    2) As the ammonia 880 solution is discussed in terms of grams, it is best to know how much 2 moles of ammonia weighs. Multiplying the 2 moles by the Mr (17) gives us 34 grams
    3) The hardest part of this question is to understand that the total ammonia 880 solution weighs 0.880 grams per cm3. As we are told 28% of this solution is ammonia, we can say that (0.880 x 0.28) grams of ammonia are present per cm3, which is 0.2464g
    4) We want 34 grams of ammonia. We know that 1cm3 of the ammonia 880 solution contains 0.2464 grams of ammonia. To work out the required volume, simply do (34/0.2464) = 138cm3 , or 0.138dm3
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    Thank you to the both of you. Passed my Pre-Test exam two days ago.
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    (Original post by Chqr)
    Thank you to the both of you. Passed my Pre-Test exam two days ago.
    Top man, congrats.

    Good luck with your chemistry career!
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    (Original post by Chqr)
    Thank you to the both of you. Passed my Pre-Test exam two days ago.
    Hey, erm I passed that test as well and received my offer today. What about you?
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    What is this Universtiy Pre-test for? I haven't heard about it before :confused:
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    (Original post by Guddi)
    Hey, erm I passed that test as well and received my offer today. What about you?
    I got an offer but I'm staying in the UK and going for Biomedical Sciences.
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    Just to say, it may not matter much now depending on your attitude towards actually learning the chemistry (as opposed to passing the exams) but the answer you've given to question 3 is certainly wrong. K = [NH3]^2 / ( [N2] * [H2]^3), feeding numbers in gets you 0.084 mol^(-2) * dm^6.
 
 
 
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