pukebox Posted October 15, 2004 Posted October 15, 2004 I have 3% of NaCl stock solution. How do I make these concentrations with only 3% of NaCl? How much distilled water do I need to dilute before I can get these concentrations? -> 0.2% -> 0.4% -> 0.6% -> 0.85% -> 1% -> 2% -> 3%
YT2095 Posted October 15, 2004 Posted October 15, 2004 well 1`stly you`ll need a common denominator we`ll use ml. the 3% you don`t need to do anything with for the 2% put in 66.6 ml of your 3% soln into a 100ml flask and top the rest up to 100ml with your water using this 2% soln pour 50ml of it into an empty 100ml flask and fill that up to 100ml with water, there`s your 1% soln and just continue like this all the way down
pukebox Posted October 15, 2004 Author Posted October 15, 2004 Hokey I'm kinda confused. It's 3% concentration of NaCl stock solution (I think). And why do I have to put in 66.6ml? Because it's 2% divided by 3%? So if I wanna find solve 0.85% it'll be 0.15% divided by 3%? Therefore I put in 5ml of my 3% solution into a 100ml flask, top the rest up to 100ml with water, pour 50ml into another 100ml flask, fill it up to 100ml and there's my 0.85%?
YT2095 Posted October 15, 2004 Posted October 15, 2004 because 66.6 ml and 33.3 ml will make 100ml of 2% soln
pukebox Posted October 15, 2004 Author Posted October 15, 2004 because 66.6 ml and 33.3 ml will make 100ml of 2% soln Okay.. but why not 50ml, or 40ml, y'know.
YT2095 Posted October 15, 2004 Posted October 15, 2004 100ml is a nice neat figure to work with since we`re talking Percent, which is based on 100 and using ml is a nice tiny amount that doesn`t require Buckets or a microscope
badchad Posted October 15, 2004 Posted October 15, 2004 I always use: C1 V1 = C2 V2 (concentration of solution 1)(volume of solution 1) = (conc. sol. 2)(vol. of sol. 2) This way you don't necessarily need a common denominator. For instance, we know solution #1 is 3%, (your stock). We also know the concentration we want (conc. #2). All you have to do is decide how much of number 2 you want. Using YT2095's 100ml, you'd set it up as: (3%) (V1) = (.2%)(100ml) Solve for v1. Remember 100 ml is the final volume. So in this case v1= 6.67. You.d put 6.67ml in 93.33 ml of water. Sound correct YT2095? (just an alternate way of doing it)
YT2095 Posted October 15, 2004 Posted October 15, 2004 sure that`s fine too at the end of the day it`s only basic maths, whether it be ml gr marbles or baked beans, subdivision is just that, you`re perfectly right also. it just beats me that a simple answer wasn`t understood and a more complex one was? LOL ))
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