• Motivanka
In summary, you should add 227.78 mL of water to 50 mL of glucose solution to get a solution with a concentration of 0,002 mol/dm3.f

#### Motivanka

Hello!
I have few questions regarding concentrations and I am really confused:
1. Task: How many cm3 of water should be added on 50 cm3 solution of glucose which has mass concentration 2 g/dm3 to get solution that has concentration 0,002 mol/dm3 ?
So what happened is that I got correct result but I think its not good procedure:

V1(H2O) =?
V2(C6H12O6)=50cm3=0,05dm3
ϒ2(C6H12O6)=2 g/dm3
C2(C6H12O6)=0,002 mol/dm3

V1⋅ϒ1=V2⋅ϒ2
V1=(V2⋅ϒ2) / ϒ1

ϒ1=?

C2= n2 / V || n2=m2 / M2
C2=m2 / (M2 ⋅ V ) || ϒ1 = C2 ⋅ M2 ⇒ This is what confused me. If I put all data above in this I'll get 0,36 g/dm3

and after that when I put data in formula V1 = (V2⋅ϒ2) / ϒ1 I'll get V1=0,277 dm3 = 277,78 cm3

AND THAT IS CORRECT RESULT ! BUT how come I can get mass concentration of water while using data for glucose this is what consfuses me ! If I haven't done it well can someone explain me or give me any idea please.
What confuses me the most is relation of different solution and solvent I can't relate them by formulas. I am ok with these tasks : How many cm3 20% solution of HCl density ∫=1,1 g/cm3 is required for prepearing 1dm3 HCl if C=2 mol/l BUT when it comes to different solution and solvent I am stuck.

Sorry for bad english its not my native language :D

The volume you got was of the final solution = 277.78 mL and hence, the water to be added is = (277.78-50)mL = 227.78mL.

There is another easy method which is called "Unitary Method" and you also don't have to remember any formulas for that

There are ## 2*0.001 ## gms glucose in 1 mL
so in 50mL there are ## 0.1 ##gms glucose

and let x mL of water be added
This is Dilution so gms of glucose will not change
so in (50+x)mL of final solution there are ## 0.1 ##gms of glucose

Now it is easily solvable as you know the concentration of the final solution is ## 2*0.000001 mols/mL ##

Thank you so much for your help :D I got an idea. When you said " the volume you got was of the final solution = 277.78 mL and hence, the water to be added is = (277.78-50)mL = 227.78mL." which means that what upset me was mass concentration of whole solution not only for water, so after finding it I should put it in this formula V=(V2⋅ϒ2)/ϒ and Ill get volume of whole solution and then detract it with 50 cm3 and get VH2O=227,78 ml. Ill take for further tasks second way. Thanks again :)