How do i calculate a 5 millimolar solution of glucose ?

Answer 1

First find the formula for glucose
Glucose (C6H12O6)
--> http://en.wikipedia.org/wiki/Glucose#Structure
Then find it's molecular weight ( by adding the atomic weight of the atoms or by looking at the reference above)
180.16
Then understand what a millimolar solution is
.001 moles / liter
then calculate
(180.16 (g / mole)) * .001 (moles / liter) * 5 = 0.9008 g / L

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Answer 2

You do not. calculate 5mmol.
It is an amount.
5 millimolar solution means that 5 thousandth of one mole of glucose is contained in a total volume of one litre. Mathematically it is written 5.0 x 10^-3 or (0.005)
Using C6H6O6 as formula for glucose then its Relative Molecular Mass is :-

C : 6 x 12 = 72
H : 6 x 1 = 6
O : 6 x 16 = 96
___
174 (Mr of Glucose)
So mass of 5 mmol is :- 0.005 x 174 = 0.87 grammes
So 0.87 g would be dissolved in distilled water and made up to a volume of one litre.
This would be a 5mmol solution of glucose.

Answer 3

5 mM is 0.005 Molar.

1 Molar is 1 mole in 1 litre so you need to know what volume you are making.

1 mole is calculated by dividing the mass by the molecular mass, which for glucose is 180.

So, 0.005 M * 1 L (1 litre volume) = 0.005 moles

0.005 moles * 180 (molecular mass) = 0.9 g

so if you add 900 milligrams of glucose to 1 litre of water you get a 5 milliMolar solution.

Answer 4

assuming you want to make 1 litre of the solution its dead easy.

Just work out the mass of 5 millimoles of glucose and dissolve it in a litre of water.



to do this you need the Mr of glucose (C6H12O6)first.
Mr = (6 x 12)+(12 X 1) + (6 x 16) = 180

then: if n = number of moles of glucose

n (moles) =mass of glucose needed in grams / Mr of glucose

so rearranged for mass (the thing you need to know):

mass (grams) = n (moles) x Mr

and remember 1 mmol = 1/1000 mol so

mass (grams) = (5/1000) x 180

mass (g) = 0.005 (moles) x 180

mass (g) = 0.90g

take 0.90g of 100% pure soild glucose and dissolve it in 1litre of pure water and that will be a 5mmolar solution of glucose.

I am a geek!

Answer 5

I basically agree with the others but when you make up 1 litre of solution, it is important that the final total volume, solute and solvent, make up 1 litre. You do not add 1 litre of water to the glucose as this will give you 1 litre plus of solution, which will not have the correct strength, it will be too dilute.

Answer 6

What has been written so far is fine, just check on the % glucose in what you use, as the standard glucose is sold as the mono-hydrate -in which case one just add 18 to the molecular weight. This is the safest product to use, as an anhydrous glucose is very hygroscopic (it absorbs water from the air very rapidly).
However the purity is seldom 100%, so if its 98% pure, just divide the weight by 98%.
I do hope the detail does not confuse, but its really all very simple.

Answer 7

Glucose, C6H12O6.
Relative molecular mass, Mr = (12x6)+(1x12)+(16x6) = 180
Mass = number of moles x Mr = (5x10^-3) x 180 = 0.9g

Answer 8

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0.0023 gm=4.39 X10^-6 moles. that quantity would =volume times molarity required 4.39times 10^-6=vol(litres)times 20X 10^ -3 =V times 2^10^-2 V(litres) =(4.39times 10-6)/(2times 10^-2) =(4.39/2) times10^(-6-(-2)) =2.195 times10^-4 litres =about 2.2 times 10^ -1 mls. =2.195 times 100 microlitres 219.5 microlitres. If I've got this wrong I'm sure I will soon get the horse laugh. Good Luck,But check my sums.

Answer 9

milimolar solution