The question is: Isotonic saline is a 0.9% (g/100ml) solution of salty water. You are asked to make up 2L of isotonic saline. How much salt should you weigh out?

Question

I got the answer of 18g with the following working:

9 ÷ 1000(ml) x 100 = 0.9
Therefore 9% of 1000ml (1L) is 9g
Multiply 9g by 2 because it is for 2L of water = 18g

Is this method of calculation suitable for a chemistry assignment or is there another way to do it?

Thanks.

Answer 1

There is nothing wrong with your answer and your math appears ok but the reasoning is out of this world. You apparently picked 9 as a starting point (and it's good that you did); however, if you had picked any other number it wouldn't have given you the correct answer. Here is a (there are several) conventional way of doing it.
0.9% solution = 0.9g/100 g solution (which is the same as substituting into the percent equation as
%NaCl = (g NaCl/100 mL)*100 = (0.90 g NaCl/100 mL)*100 =0.9% NaCl. Now if we want 2.00 L, just multiply by the factor of what you want over what you have; i.e., (0.90g NaCl) x (2,000 mL/100 mL) = 18 g