How do I do this glucose dilution problem?

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SUMMARY

The discussion focuses on preparing a 25 ml solution of 2.5% glucose from a stock solution of 1.75 M glucose. The key calculation involves using the equation c1v1 = c2v2, where c1 is the concentration of the stock solution (315 g/L), c2 is the desired concentration (0.625 g in 25 ml), and v1 is the volume of the stock solution needed. The final result indicates that 1.98 ml of the stock solution should be mixed with 23.02 ml of water to achieve the desired concentration without needing the density of the glucose solution.

PREREQUISITES
  • Understanding of molarity and mass/volume percentage calculations
  • Familiarity with the equation c1v1 = c2v2 for dilution problems
  • Knowledge of molecular weight calculations, specifically for glucose (180 g/mol)
  • Basic laboratory skills for preparing solutions
NEXT STEPS
  • Study the concept of molarity and how to convert between different concentration units
  • Learn about preparing dilutions in a laboratory setting
  • Explore the implications of solution density in various chemical contexts
  • Review common laboratory practices for measuring and mixing solutions accurately
USEFUL FOR

This discussion is beneficial for chemistry students, laboratory technicians, and anyone involved in solution preparation and dilution calculations.

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Homework Statement



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

Homework Equations


c1v2 = c2v2

molecular mass glucose = 180g/mol

The Attempt at a Solution



not sure if I am doing this right:

2.5 % m/v

2.5g in 100ml water (i got help with this definition, but i still don't 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:
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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! :biggrin:. 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.
 

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