1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Tasks with concentration

  1. Jun 2, 2016 #1
    I have few questions regarding concentrations and I am really confused:
    1. Task: How many cm3 of water should be added on 50 cm3 solution of glucose which has mass concentration 2 g/dm3 to get solution that has concentration 0,002 mol/dm3 ?
    So what happened is that I got correct result but I think its not good procedure:

    V1(H2O) =?
    ϒ2(C6H12O6)=2 g/dm3
    C2(C6H12O6)=0,002 mol/dm3

    V1=(V2⋅ϒ2) / ϒ1


    C2= n2 / V || n2=m2 / M2
    C2=m2 / (M2 ⋅ V ) || ϒ1 = C2 ⋅ M2 ⇒ This is what confused me. If I put all data above in this I'll get 0,36 g/dm3

    and after that when I put data in formula V1 = (V2⋅ϒ2) / ϒ1 I'll get V1=0,277 dm3 = 277,78 cm3

    AND THAT IS CORRECT RESULT ! BUT how come I can get mass concentration of water while using data for glucose this is what consfuses me ! If I haven't done it well can someone explain me or give me any idea please.
    What confuses me the most is relation of different solution and solvent I cant relate them by formulas. I am ok with these tasks : How many cm3 20% solution of HCl density ∫=1,1 g/cm3 is required for prepearing 1dm3 HCl if C=2 mol/l BUT when it comes to different solution and solvent I am stuck.

    Sorry for bad english its not my native language :D

    Thanks in advance !!!
  2. jcsd
  3. Jun 2, 2016 #2
    The volume you got was of the final solution = 277.78 mL and hence, the water to be added is = (277.78-50)mL = 227.78mL.

    There is another easy method which is called "Unitary Method" and you also dont have to remember any formulas for that :wink:

    There are ## 2*0.001 ## gms glucose in 1 mL
    so in 50mL there are ## 0.1 ##gms glucose

    and let x mL of water be added
    This is Dilution so gms of glucose will not change
    so in (50+x)mL of final solution there are ## 0.1 ##gms of glucose

    Now it is easily solvable as you know the concentration of the final solution is ## 2*0.000001 mols/mL ##
  4. Jun 2, 2016 #3
    Thank you so much for your help :D I got an idea. When you said " the volume you got was of the final solution = 277.78 mL and hence, the water to be added is = (277.78-50)mL = 227.78mL." which means that what upset me was mass concentration of whole solution not only for water, so after finding it I should put it in this formula V=(V2⋅ϒ2)/ϒ and Ill get volume of whole solution and then detract it with 50 cm3 and get VH2O=227,78 ml. Ill take for further tasks second way. Thanks again :)

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted