Chemistry
posted by Emma on .
I need to know how to dilute a 1L IV bag of 0.45% NaCl with 3.5% NaCl water to make it a drinkable 0.9% NaCl

The question is written in a very odd way. There is no need add more NaCl to make the 0.45% drinkable as it is already drinkable.
However, in the spirit of the question, we need to add v ml of the 3.5% solution (=sea water!) to 1 liter and end up with 0.9%.NaCl. So the starting solution contains 4.5 g NaCl (I am assuming that the % is w/volume) we need to end up with
(1000+0.0045)+(vx0.035)g salt
in 1000+v ml
and these two divided x 100/1 = 0.9%
so
[(1000+0.0045)+(vx0.035)]x100/(1000+v)=0.9%
solve for v
I got 4.5 = 8.965V+0.009V^2
if we assume that v is small
I got v=8.965/4.5 = 1.99 ml
But check my maths.