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?
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.
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
Your method of calculation is correct for determining the amount of salt needed to make up 2L of isotonic saline. However, there are a couple of additional steps you can take to further validate your answer.
First, convert the 2L to milliliters (ml) since your concentration is given in terms of grams per 100ml. 2L is equal to 2000ml.
Then, multiply the concentration (0.9g/100ml) by the volume (2000ml) and divide by 100 to find the amount of salt required:
(0.9g/100ml) x 2000ml ÷ 100 = 18g.
So, your answer of 18g is indeed correct.
This method is suitable for a chemistry assignment as it takes into account the concentration of the saline solution and the desired volume. It demonstrates a clear understanding of how to calculate the amount of solute needed to prepare a certain volume and concentration solution.
It is always important to double-check your calculations and consider the units involved to ensure accurate results.