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I have few questions regarding concentrations and I am really confused:

**1. Task: How many cm**

^{3}of water should be added on 50 cm^{3}solution of glucose which has mass concentration 2 g/dm^{3}to get solution that has concentration 0,002 mol/dm^{3}?So what happened is that I got correct result but I think

**its not good procedure**:

V

_{1(H2O)}=?

V

_{2(C6H12O6)}=50cm

^{3}=0,05dm

^{3}

ϒ

_{2(C6H12O6)}=2 g/dm

^{3}

__C__

_{2(C6H12O6)}=0,002 mol/dm^{3}V

_{1}⋅ϒ

_{1}=V

_{2}⋅ϒ

_{2}⇒

V

_{1}=(V

_{2}⋅ϒ

_{2}) / ϒ

_{1}

ϒ

_{1}=?

C

_{2}= n

_{2}/ V || n

_{2}=m

_{2}/ M

_{2}⇒

C

_{2}=m

_{2}/ (M

_{2}⋅ V ) || ϒ

_{1}= C

_{2}⋅ M

_{2}⇒ This is what confused me. If I put all data above in this I'll get 0,36 g/dm

^{3}

and after that when I put data in formula V

_{1}= (V

_{2}⋅ϒ

_{2}) / ϒ

_{1}I'll get V

_{1}=0,277 dm

^{3}= 277,78 cm

^{3}

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 cant relate them by formulas. I am ok with these tasks : How many cm

^{3}20% solution of HCl density ∫=1,1 g/cm

^{3}is required for prepearing 1dm

^{3}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 !!!