# How do I do this glucose dilution problem?

1. Sep 16, 2011

### vande060

1. The problem statement, all variables and given/known data

how do you prepare a solution for 25ml of 2.5% glucose form a stock of 1.75M glucose

i don't get how you would do this without the density of 2.5% glucose solution

2. Relevant equations
c1v2 = c2v2

molecular mass glucose = 180g/mol

3. The attempt at a solution

not sure if im doing this right:

2.5 % m/v

2.5g in 100ml water (i got help with this definition, but i still dont understand how you can say this. is it because 100ml = 100g water and 2.5g/100g = .025 of 2.5 %?)

25g in 1l

so i need a solution that has 25g/liter

----------------

1.75M * 180g/mol = 315g/l

c1v2 = c2v2

315(v1) = 25*25

v1 = 1.98

with sig figs = 2ml of the original solution needed

Last edited: Sep 16, 2011
2. Sep 16, 2011

### epenguin

You say you don't know how to do it but you have almost done it. Not completely checked but it looks right.

Fact is "2.5% glucose" is a bit ambiguous. Therefore anyone who wants it shouldn't mind a small error! . But from the rest I think we can agree it's 2.5g glucose per 100ml solution. Solution, not water.

But OK if you want not 100ml but 25 ml of that, it takes a quarter that is, er, 2.5/4 g = 0.625 g

So of a 315 g/l solution 0.625g is contained in 0.625/315 l = 0.00198 l = 1.98 ml as you found.

So actually to get this in a final 25 ml you need to mix your 1.98 ml with (25 - 1.98) = 23.02 ml of water. You do not need to know the density of anything. More often than not you won't and when you do they will tell you, so when they don't tell you you will mostly be right to think "Aha, that means for this problem I don't need to know density, so... need thinks smiley.