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### G.11 / Volumes for Solutions

Posted: Tue Apr 10, 2018 4:18 pm
So an example of this came up in G 11: A medical researcher investigating the properties of intravenous solutions prepared a solution containing 0.278 m C6H12O6 (glucose). What volume of solution should the researcher use to provide 4.50 mmol C6H12O6?

For this problem I understand that 0.278 is my M initial and 4.50 mmol (so .0045 mol) is my M final. For questions like these though do you always assume your V final is going to be 1 liter unless otherwise stated and you are finding the V initial to give you 1 liter of .0045 solution?

### Re: G.11 / Volumes for Solutions

Posted: Tue Apr 10, 2018 4:30 pm
Volume is equal to moles/concentration, so you would do 4.5x10^-3 mols divided by 0.278M.
Hope this helps

### Re: G.11 / Volumes for Solutions

Posted: Tue Apr 10, 2018 5:05 pm
For this question you don't need to know the final volume since you are given the final mole which is 0.0045 mol C6H12O6 so you can set up your equation as 0.728 M (M=mol x L^-1) x V = 0.0045 mol and to get the initial volume you would divide both sides by 0.728 mol x L^-1 which equals 6.18 x 10^-3 L.