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Topic: Chemistry Help on the subject of: % concentration, mg/mL, osmoles, osmolarity  (Read 17484 times)

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Offline boostar

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Hi all, well basically, I've been set these quizzes that I am meant to do, and I need 12/15 to pass... which isn't happening. I'm having so much trouble and I really really really need some help. Well I'll post some questions that I hope you guys can help me understand.

1. Magnesium sulphate (MgSO4.7H2O) is formulated in an ampoule (container) containing 2.47g/5mL (MW = 246). Assuming complete dissociation, how many milliosmoles are present in one ampoule?
Give answer to 3 significant figures

ok so moles=m/MM therefore there is 2.47/246 =0.01004065moles. Milliosmoles: therefore find concentration M=n/v =2.008
mOsmoles/L= 2*1000* 2.008 =4016
Significant figures means 4020mOsmoles/L... is this correct?

2. 17. An colon lavage solution contains the following in each 4L of solution:
sodium chloride                                5.84g
potassium chloride                            2.98g
sodium bicarbonate                          6.72g
sodium  sulphate                               22.72g
polyethylene glycol (MW=3350)       240g
Assuming the above electrolytes dissociate completely (& no dissociation of polyethylene glycol) and the following molecular weights:  NaCl  = 58.4, KCl = 74.6, NaCO3= 84.0, Na2SO4= 142.
Calculate the millimoles/L of chloride ions in the lavage solution?
Express answer to 3 significant figures

(Edit: Ok time to put up working:)

Millimoles/L of Cl ions
Find moles of NaCl and KCl as these have the Cl ions.
5.84/58.4 = 0.1 moles
2.98/74.6=0.03994 moles therefore there are:
0.13994moles of Cl in solution
Molarity = 0.13994/4 = 0.034985M
*1000 = 35.0mmol

3. A wart paint contains Lactic acid 20g, Salicylic acid 20g and 80g of flexible collodion.
What is concentration of salicylic acid in % w/w?
Express answer to 2 significant figures

I put 17 for this but apparently it is not right.
eg. 20+20+80 = 120

120/100 =1.2
20/1.2 = 16.67 than round to 2 sig figures.

4. 10. How many milliosmoles are in a 200 mL infusion of 5 % w/v glucose?
(MW glucose = 180, a non electrolyte)
Round your answer to 3 significant figures.
(I got 278 but apparently this was wrong :S)

5%w/v means 5g/100mL
therefore 10/180=0.05555
Molarity: 0.055555/0.2=0.277777
mOsmol/L= 0.27777 * 1000
=278

5. 2. An electrolyte solution contains 222 mg of sodium acetate (Na C2H3O2 = 82) and 15 mg of magnesium chloride (MW MgCl2= 95.0) in 100 mL of solution. Assuming 80% dissociation, calculate the number of milliosmoles/L in this solution?
Express answer rounded to 3 significant figures.

222*10^-3 / 82 =0.002707317
15*10^-3/95 =0.000157894
Add these than divide by 0.1 for Molarity = 0.02865211
mOsmoles/L= 0.028652117 * 5 * 1000 * .80
=115mOsmole/L

6. 12. An colon lavage solution contains the following in each 4L of solution:
sodium chloride                                5.84g
potassium chloride                            2.98g
sodium bicarbonate                          6.72g
sodium  sulphate                               22.72g
polyethylene glycol (MW=3350)       240g
Assuming the above electrolytes dissociate completely (& no dissociation of polyethylene glycol) and the following molecular weights:  NaCl  = 58.4, KCl = 74.6, NaCO3= 84.0, Na2SO4= 142.
What is the osmolarity (milliosmols/L) of this solution?
Express answer to 3 significant figures

Milliosmols:
Number of ions in solution = 11
find mols of each than add them = 0.4454moles
Find molarity= 0.4454/4 =0.111M

mOsmol/L = 0.111*1000*11
=1221
Sig figures = 1220


(Edit: putting up working)



« Last Edit: March 27, 2008, 07:10:44 AM by boostar »

Offline AWK

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4020 - four significant digits
35 - two significnt digits
AWK

Offline ARGOS++

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Dear Boostar;

Don’t care for the Sig. Digits as long as your Calculations are so far from the real asked Results!

To 1.     )  It is not a Concentration asked, but the Content per Ampoule in Milli-Osmoles!
To 2.17)  The value looks ok, but you forgot the Units! 
              Only 35 can mean all thinkable, also Potatoes!  (Now take care for Sig. Digits.)
To 3.     )  The same as for 2.17).
To 4.10)  Result is wrong.  The Result is already done in Line 2.
              But convert it to the correct Units and apply the Sig. Digits.
To 5.02)  Result is wrong.  Both Salts dissociate not to the same Number of Ions!
To 6.09)  Why is the Result different to the one of  2.17)?


Good Luck!
                    ARGOS++


Offline boostar

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To 1.     )  It is not a Concentration asked, but the Content per Ampoule in Milli-Osmoles!

Ok, so I found the mOsmol/L, but we need the mOsmol for 5mL, so 1000/5 = 200 therefore I should divide 4016 by 200 to get 20.1 mOmols in 5 mL... correct?

To 4.10)  Result is wrong.  The Result is already done in Line 2.
              But convert it to the correct Units and apply the Sig. Digits.

I'm not really sure what you mean there, but is it because I needed to find the mOsmol per 200mL?

"5%w/v means 5g/100mL
therefore 10/180=0.05555
Molarity: 0.055555/0.2=0.277777
mOsmol/L= 0.27777 * 1000
=278"

So than Because this is mOsmol per 1000mL, I just divide it by five to get mOsmol of 200mL = 55.56 mOsmol?

To 5.02)  Result is wrong.  Both Salts dissociate not to the same Number of Ions!

Sorry? Could somebody clarify what he means?
           
            "5. 2. An electrolyte solution contains 222 mg of sodium acetate (Na C2H3O2 = 82) and 15 mg of magnesium chloride (MW MgCl2= 95.0) in 100 mL of solution."

I thought that 5 ions would dissociate in solution because Na+, C2H3O2-, Cl-, Cl- and Mg+ would dissociate?


To 6.09)  Why is the Result different to the one of  2.17)?

Because it asks for "the osmolarity (milliosmols/L) of this solution," while the other asks for the "millimoles/L of chloride ions in the lavage solution."

Thanks for your posts AWK and ARGOS++ :)

Offline ARGOS++

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Dear Boostar;

Please!!  Don’t make the Corrections of your Calculations in the Original Posting!
It is very hard to see what you have really changed in the meantime!


To 1.     )  Now the Result is correct!
              Your first calculation: 
              2.47 / 246  = 10.04065 mmoles is already the content in 5 mL!
              But now you have to multiply it with the Number of Ions (= 2)!

To 2.17)
Calculate the millimoles/L of chloride ions in the lavage solution?
               Unit still wrong!   Sig. Digits ok.

To 3.     )  No changes detect. Unit still missing.  (Now take care for Sig. Digits.)

To 4.10)   Now the Result is correct!   But you can do it easier, as in your line 2!
               5% w/v of 200 mL = 10.00 g.   
               And 10.00 g / 180 g/m = 55.5 mmoles (per 200 mL).   That’s all!

To 5.02) a.) How many Ions will be generated from NaCl?   (And how many mMoles?)
             b.) How many Ions will be generated from MgCl2? (And how many mMoles?)

To 6.12) a.) How many Ions will be generated from NaCl? (And how many mMoles?)
             b.) How many Ions will be generated from KCl?   (And how many mMoles?)
             c.) How many Ions will be generated from NaHCO3?  (And how many mMoles?)
             d.) How many Ions will be generated from Na2SO4?   (And how many mMoles?)
             e.) How many Particles are generated from PEG?  (And how many mMoles?)
             f.) And that makes in total?

I hope to have been of help to you.

Good Luck!
                    ARGOS++

« Last Edit: March 27, 2008, 10:36:57 PM by ARGOS++ »

Offline boostar

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You've been an unbelievably big help  :-*... so much in fact that I feel bad asking for more  :-[

(Btw I haven't made any changes to my original post: I'm only copying and pasting my original working)

To 5.02) a.) How many Ions will be generated from NaCl?   (And how many mMoles?)
             b.) How many Ions will be generated from MgCl2? (And how many mMoles?)

Here is the question:
            "5. 2. An electrolyte solution contains 222 mg of sodium acetate (Na C2H3O2 = 82) and 15 mg of magnesium chloride (MW MgCl2= 95.0) in 100 mL of solution. Assuming 80% dissociation, calculate the number of milliosmoles/L in this solution?"

Ok So 2 ions will be dissociated from Sodium Acetate - Na, and C2H3O2. And from MgCl2, there will be three ions generated... Mg and 2 Cl ions. Is this correct? NOw when you ask for how many mMoles, what do you mean? Do you mean the mMoles of the sample substance in solution? I'll show some new working than:

Find moles of Sodium Acetate: 222*10-3 / 82 = 0.002707... therefore this *2 is how many moles of Na and C2H3O2 dissociate in solution.
Find moles of MgCl2: 15*10-3 / 95 = 0.0001157, therefore there will be 0.0003157 moles of Cl and 0.000157 moles of Mg that dissociate into solution???

So total moles = 0.002707*2 + 0.0001557*3 = 0.00588 moles

Find molarity: 0.00588/0.1 = 0.0588

Find mOsmoles/L= 0.0588*5 (total number of moles of ions that dissociate in solution???) * 1000
=294mOsmol/L

Hmmmm? I'll have a go at number 6 once I understand this one.

Offline ARGOS++

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Dear Boostar;

To 5.02)  Why are you multiplying the intermediate result by 5 ??

You have already done the Multiplication for the individual Number of Ions; you did it this times correct, and that’s enough!
You have now only to calculate for 80% Dissociation: 58.89 mMoles/L * 0.8 =  47.1 (mMoles/L) = Milli-OsMolarity.
And that’s again all!

I’m sure, if you do 6.12) the same/similar way, you will get the correct Result.

Good Luck!
                    ARGOS++

Offline boostar

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Dear Boostar;

To 5.02)  Why are you multiplying the intermediate result by 5 ??

You have already done the Multiplication for the individual Number of Ions; you did it this times correct, and that’s enough!

Sorry, when did I do the multiplication for the individual ions?

My understanding is that:

mOsmol/L = Molarity * number of ions that dissociate in solution

So I found the total number of moles and divided by volume to get molarity, and than to get the osmolarity, I multiplied by the number of ions that dissociated. Where have I gone wrong?



Offline ARGOS++

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Dear Boostar;

You have only the lines you wrote full explained and correctly to reread:
So total moles = 0.002707*2 + 0.0001557*3 = 0.00588 moles
Find molarity: 0.00588/0.1 = 0.0588
==>  Total moles of Ions per 100 mL = 
           0.0027074  mole NaAcetate * 2 Ions  =  0.0054147 mole Ions / 0.1 L    +
           0.0001574  mole MgCl2       * 3 Ions  =  0.0004722 mole Ions / 0.1 L
                                                     Total  =  0.0058869 mole Ions / 0.1 L
                                                              =  58.869  mmole Ions / L

And now continue better then I did late last night (4:32 AM):
  80 % Dissociation =  80 % Ions + 10 % “Particles” = 90 % Species    ==>
            Osmolarity =  58.869 mmoles / L * 0.9  =  52.98  mOsmoles / L

And that’s now all.   (Sorry for my mistake!)


Good Luck!
                    ARGOS++

Offline boostar

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Once again thankyou for your reply, it's really appreciated.

And now continue better then I did late last night (4:32 AM):
  80 % Dissociation =  80 % Ions + 10 % “Particles” = 90 % Species    ==>
            Osmolarity =  58.869 mmoles / L * 0.9  =  52.98  mOsmoles / L

Ok, I'm with you until I get to the above bit. So we have 80% dissociation, but what is with the 10% particles? And why can you add it to the 80% dissociation?

Offline ARGOS++

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Dear Boostar;

That should be quite simple!
Try to think by yourself: 
  • If only 80% of the molecules are dissociated, where and in which form are the remaining molecules?
  • Do they have no participation on the Osmolarity?
  • And how much is 20% not dissociated Ions in %-age of molecules with respect of 100% Salt?

Good Luck!
                    ARGOS++


Offline boostar

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    Dear Boostar;
    • If only 80% of the molecules are dissociated, where and in which form are the remaining molecules?
    • Do they have no participation on the Osmolarity?
    • And how much is 20% not dissociated Ions in %-age of molecules with respect of 100% Salt?

    Ok 80% of the molecules dissociate: which means that the remaining molecule are on the bottom still int he compound, which also means they cannot participate in the osmolarity. your last question is harder to answer though, I'm sorry but I don't understand the wording :( So 20% undissociated ions means that 20% of salt will remain undissociated?


    Offline ARGOS++

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    Dear Boostar;

    Now your “Picture” and your Answers are wrong!

    Because:
    2. 17. An colon lavage solution contains the following in each 4L .. :
    Also un-dissociated Molecules can be solvated “exactly” in the same kind as Ions are.

    Let’s draw a simpler Picture, which you may be able to catch:
    -  In case of 100%  Dissociation:
             50 molecules (z = 2)                         --------> =  100 Ions.
    -  In case of   80%   Dissociation:
             50 molecules (z = 2)    *  0.8             --------> =    80 Ions      =  80 % Species.
             50 molecules (z = 1)    * (1.0 - 0.8 )   --------> =   10  Particles or Molecules.
                                                                 Total       =    90 Species  ~ 90 % Species.


    Good Luck!
                        ARGOS++

    Offline boostar

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     ???  :'(
    Let’s draw a simpler Picture, which you may be able to catch:
    -  In case of 100%  Dissociation:
             50 molecules (z = 2)                         --------> =  100 Ions.

    Um what's z? Well 50 * 2 = 100 and 100% dissociation so there are 100 ions.. and example of this would be NaCl right?

    -  In case of   80%   Dissociation:
             50 molecules (z = 2)    *  0.8             --------> =    80 Ions      =  80 % Species.
             50 molecules (z = 1)    * (1.0 - 0.8 )   --------> =    10  Particles or Molecules.

    Ok so 50 * 2 = 100 in total but only 80% dissociates therefore there are 80 ions

    Now the next bit is confusing... you are saying that undissociated molecules also affect osmolarity?

    50 molecules (z = 1)    * (1.0 - 0.8 )   --------> =    10  Particles or Molecules.

    So z=2 was used previously because 2 types of ions dissociated from the compound. Here, z=1 because we are talking about the compound? Let me try to illustrate my understanding:

    1) Lets pretend 50 moles of NaCl dissociates 80%. Therefore:
               
    Because it dissociates into Na and Cl z=2, and therefore 50*2*0.8 = 80 ions= 80% of species

    Now we still have the Original NaCl compound, so z=1 because it is one big particle? So therefore if we have one mole of NaCl, it must mean that: 1-0.8 = 0.2 = 20% undissociated? Therefore this 20%, or 10 particles are also included in the osmolarity as particles present in the solution? Therefore there are 90 species in total. Hopefully this is correct?

    Ok Lets try something harder:

    2) Lets pretend 50moles of MgCl2 dissociates 80% in solution

    MgCl2 (z=3): 0.8 * 3 * 50 = 120
    MgCl2 where z=1 because we count it as the large compound: (1-0.8) * 1 * 50 = 10 particles of MgCl2 remain undissociated, but still participate in the osmolarity?

    Therefore 10 + 120 species in total are present???







    Offline ARGOS++

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    Dear Boostar;

    You’re absolutely correct!      (But if you name the Ions you should add the charge too.)

    Your understanding of z is also correct, and also the 10 + 120 is correct calculated.
    And how ‘many’ Percent instead of 90% are those?
    Do you see already the general formula for calculating that for different z?

    And then you can calculate how large my Error would be for your calculation, and then you know also why I dropped the Difference for MgCl2 to hold it as simple as possible (- as I told you at the Begin).

    Additionally you should keep in mind, that “in real” not both Salts would be identically affected by this Dissociation Ratio, but no value is given for that in your Question.

    But I’m happy, that your understanding has started so effectively!

    Good Luck!
                        ARGOS++


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