can you prove that 5% dextrose water is mathematically equivalent to .9% normal saline?

Question

plz give refrences

Answer 1

I guess you are reffering to osmotic pressure since these solutions are commonly used for intravenous infusion.

Dextrose is D-glucose monohydrate so it has MW=198
Part of your 5 gr (per 100ml of solution) is water, so you need to find out how much glucose you actually have in there.
You have (MWglucose / MWdextrose)*5 =(180/198)*5 grams
and therefore the molarity is
C= mass/(MW*V)= (180/190)*5 /(180*0.1) = 0.263 M

Therefore the osmotic pressure is

π1= CRT= 0.263RT

Saline is NaCl with MW=58.5
Thus C=0.9/(58.5*0.1)= 0.154 M

Since NaCl is an electrolyte
π2= iCRT and i=2 for NaCl
so π2=2*0.154*RT=0.308RT

π1/π2= 0.263RT/0.308RT= 0.85

If the two solutions were exactly isotonic the ratio would have been 1. I guess it is still close enough to perform the infusions without any problem. Actually the 0.9% saline is slightly hypertonic compared to blood:
http://en.wikipedia.org/wiki/Saline_(medicine)

Answer 2

not equal:

.9% NaCl = .154g/L

5.%dextrose = 45g/L

to be equal:

.003422222L of 5%dextrose times 45g / L = .9%NaCl of .154g/L

Answer 3

i don't think this question makes much sense.
What quantity should be equal ?
Do you mean the volume?
And.. in the percentage, do you mean mol percentage or mass percentage?

I just don't get you :(