Hello, I have a question regarding dilution:
If i need to dilute 50% dextrose into 20mg of a 2.5% solution, how would i do that?
If i need to dilute 50% dextrose into 20mg of a 2.5% solution, how would i do that?
First, convert your %solution into a Molarity: [(%solution)*10]/FW
This answer gets you g/L
Since Dextrose (or d-glucose) is C6H12O6, it has a formula weight (FW) of 180.16 using your periodic table.
= 0.02775M (keep the sig digits for now)
Do the same for a 2.5% solution: you get 0.00139M
Finally, convert 20mg to of 0.00139M C6H12O6 into liters, and use your friend M1V1 = M2V2 (M is molarity and V is volume).
I am three years removed from all this, so double check your and my work. I am assuming this "explanation" is better nothing. Especially check units, and that is always.
Hello, starcherub08!
Since we will be adding pure water, consider the amount of water at each stage.If i need to dilute 50% dextrose into 20mg of a 2.5% solution,
how would i do that?
Let = amount of water to be added.
Then = amount of the 50% solution.
We start with mgs of solution which is 50% water.
. . It contains: . mgs of water.
We add mgs of pure water.
. . It contains (of course): . mgs of water.
The mixture contains: . mgs of water. .[1]
Consider the mixture again.
. . It will be mgs of solution which is 97.5% water.
The mixture contains: . mgs of water. .[2]
We just described the final amount of water in two ways.
There is our equation: [1] = [2] . . . .
. . Got it?