Question G.11 (Sixth Edition)

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Steve Magana 2I
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Joined: Fri Sep 28, 2018 12:24 am

Question G.11 (Sixth Edition)

Postby Steve Magana 2I » Wed Oct 10, 2018 11:11 pm

Question: 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?

My answer is not matching the correct one, so I was hoping I can get some guidance to see where I went wrong on the problem. Thank you!

Chris Freking 2G
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Joined: Fri Sep 28, 2018 12:29 am

Re: Question G.11 (Sixth Edition)

Postby Chris Freking 2G » Wed Oct 10, 2018 11:42 pm

Use the formula M = n/V. In the context of this problem, solve for V.

V = n/M = (4.50 * 10^-3 mol)/(0.278 mol*L^-1) = 1.62 * 10^-2 L, or 16.2 mL

Kim Tran 1J
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Joined: Fri Sep 28, 2018 12:24 am

Re: Question G.11 (Sixth Edition)

Postby Kim Tran 1J » Wed Oct 10, 2018 11:45 pm

The question provides you with the molarity and moles of solute, and asks you to find the volume.
So you will be using the formula for molarity to solve for the volume: rearrange M=n/V to V=n/M to solve for the volume.
M= 0.278 M (or 0.278 mol/L)
n= 0.00450 mol (the question gives you 4.50 mmol but you must convert from mmol to mol. 1 mol = 1000 mmol, hence 4.50 mmol*(1 mol/1000 mmol) = 0.00450 mol)

Now you can plug these values into V=n/M:
V= 0.00450 mol/(0.278 mol/L)
V=0.0162 L or 1.62x10^-2 L
The solution manual further converts L to mL so 1.62x10^-2 L *(1000 mL/1 L) = 16.2 mL
Technically the question never specified or asked for you to convert to mL, so you would still be right if you leave your answer in L.

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