[Excelsius]

I am trying to figure out how I can calculate the error propagation when I am making dilutions. For example, let's say I have 200mL (5M NaCl, 10M KCl) 10x concentrated solution. Let's say the error is 200+-0.5mL. So to make a normal 1x solution I would add the 200mL to 1800mL water. I understand that there is an ADDITIVE error here where I take the square root of the squared errors for each flask to come up with the error, but I don't understand how I can calculate the multiplicative error. The error propagation must be different depending on the concentration of the solution I have, like 10x, 100x, 1000x, and it should go higher with higher concentrations. Or maybe I'm missing something? A clarification would be most appreciated.

[Excelsius]

Ok, I am going to simplify hoping someone can shed some light.

Let's say I have a 10±0.1mL concentrated solution that I am going to dilute 1:10 with pure water to get my final solution. Assuming the volume of the water added (90mL) is exact, what would be the error in the final solution?

Now how about the exact same scenario, but this time we dilute the concentrated solution with 1:100 water (990mL water)?

According to the rules of error propagation, it seems that the error in the first case should be 0.1x10= ±1mL and for the second case should be 0.1x100= ±10mL for the final solution. I am not sure though that this method of calculation is correct. Note that in the case of 100x dilution the error of ±10mL is actually 100% error compared to the concentrated solution of 10mL.