Tasks with concentration

[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:

V_1(H2O) =?
V_2(C6H12O6)=50cm³=0,05dm³
ϒ_2(C6H12O6)=2 g/dm³
C_2(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

Thanks in advance !

 

[Sahil Kukreja]

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 $2*0.000001 mols/mL$

 

[Motivanka]

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 V_H2O=227,78 ml. Ill take for further tasks second way. Thanks again :)

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