Question

What Molarity of NaCl would give a 300mOsm solution so as to be isotonic to ECF and ICF Set up and solve for X: (cross multiply) 300 mOsm X 2000 mOsmTM Try this on your own, then read on. Show work here: You should have an answer of 0.15 M. If you don't: please ask for help. You need to be able to get that calculation for the next part. Now find the number of grams per liter needed to make such a solution, using the molecular weight of NaCl (see previous section for this) and the molarity of 0.15 M. Take this number and divide by 1000 (this gives grams per milliliter (g/ml)). Multiple that by 100 (this give the grams per 100 ml or the % of solute in 100 ml). Review previous section on expressing concentration of a solution as a percent solution if you're not sure why you're doing this. This calculation gives the % of NaCl in a 100 Ml solution with a molarity of 0.15 M. This percent of NaCl is isotonic to the extracellular fluid. Using the same method: calculate the isotonic percent solution for glucose, KCl and for NazCO- figure out the molarity needed for a 300 mOsm solution of each, then use the molecular weights to calculate grams/liter. Finally, figure out the % in 100 ml of solvent, then you know the amounts needed for isotonic solution of each of these solutes. The molecular weight for glucose is in your Molarity lab work above. Na:CO, dissociates into three particles, one of CO, , and 2 of Na'. The molecular weight for Na2CO3 is 105.5 g/mol. The molecular weight for KCl is 74.5 g/mol, and it dissociates into 2 particles like NaCl Glucose:

Answer 1

To find the molarity of NaCl,the following cross multiplication is done as per the given data in the question

(300mOsm/2000mOsm)=(x/1M)

Where x=Molarity of NaCl solution

After doing cross multiplication we will have

x=300mOsm*1M/2000mOsm

=300M/2000

=0.15M

Since ,

Molarity

=(No. Of moles of Nacl/Volume of NaCl solution in Litre)

So,No. of moles of NaCl requred to make 1 Litre of 0.15 M NaCl =Molarity of NaCl*1 L NaCl solution

   =0.15M*1L

=0.15 moles of NaCl

Mass of 1 mole of NaCl is=23 g Na+35.5 g Cl

=58.5g

Mass of 0.15 mole of NaCl=58.5g*0.15=8.77g

Therefore 8.77 g of NaCl should be dissolved in 1 L solution to produce 0.15 M NaCl solution

Converting the concentration into g/ml

=8.77g/1 L

=8.77g/1000 ml (As 1 L=1000 ml)

=0.00877 g/ml

Again

1 ml of solution contains =0.00877 g NaCl

100 ml solution contains=0.0087*100=0.87 g NaCl

% of NaCl solute in 100 ml solution=0.87%