To 1. ) It is not a Concentration asked, but the Content per Ampoule in Milli-Osmoles!
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)?
Calculate the millimoles/L of chloride ions in the lavage solution?Unit still wrong! Sig. Digits ok.
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?)
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!
So total moles = 0.002707*2 + 0.0001557*3 = 0.00588 moles==> Total moles of Ions per 100 mL =
Find molarity: 0.00588/0.1 = 0.0588
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
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?
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.
50 molecules (z = 1) * (1.0 - 0.8 ) --------> = 10 Particles or Molecules.
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? z?
- 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.