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.
$\displaystyle \frac{(50g/100ml)*10}{180.16 g/L}$ = 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 $\displaystyle x$ = amount of water to be added.
Then $\displaystyle 20-x$ = amount of the 50% solution.
We start with $\displaystyle 20-x$ mgs of solution which is 50% water.
. . It contains: .$\displaystyle 0.5(20-x)$ mgs of water.
We add $\displaystyle x$ mgs of pure water.
. . It contains (of course): .$\displaystyle x$ mgs of water.
The mixture contains: .$\displaystyle \boxed{0.5(20-x) + x}$ mgs of water. .[1]
Consider the mixture again.
. . It will be $\displaystyle 20$ mgs of solution which is 97.5% water.
The mixture contains: .$\displaystyle 0.975(20)\:=\:\boxed{19.5}$ mgs of water. .[2]
We just described the final amount of water in two ways.
There is our equation: [1] = [2] . . . . $\displaystyle {\bf0.5(20-x) + x \:=\:19.5}$
. . Got it?