# dilute concentration

• May 16th 2008, 03:20 AM
starcherub08
dilute concentration
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?
• May 16th 2008, 04:07 AM
abender
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

Since Dextrose (or d-glucose) is C6H12O6, it has a formula weight (FW) of 180.16 using your periodic table.

$\frac{(50g/100ml)*10}{180.16 g/L}$ = 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.
• May 16th 2008, 04:14 AM
Soroban
Hello, starcherub08!

Quote:

If i need to dilute 50% dextrose into 20mg of a 2.5% solution,
how would i do that?

Since we will be adding pure water, consider the amount of water at each stage.

Let $x$ = amount of water to be added.
Then $20-x$ = amount of the 50% solution.

We start with $20-x$ mgs of solution which is 50% water.
. . It contains: . $0.5(20-x)$ mgs of water.

We add $x$ mgs of pure water.
. . It contains (of course): . $x$ mgs of water.

The mixture contains: . $\boxed{0.5(20-x) + x}$ mgs of water. .[1]

Consider the mixture again.
. . It will be $20$ mgs of solution which is 97.5% water.
The mixture contains: . $0.975(20)\:=\:\boxed{19.5}$ mgs of water. .[2]

We just described the final amount of water in two ways.

There is our equation: [1] = [2] . . . . ${\bf0.5(20-x) + x \:=\:19.5}$

. . Got it?