Fundamental G.3 Concentration

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Amy Luu 2G
Posts: 105
Joined: Wed Sep 18, 2019 12:19 am

Fundamental G.3 Concentration

Postby Amy Luu 2G » Wed Oct 02, 2019 10:08 pm

For example problem G.3A it asks:
What volume of 1.25x10^-3 M C6H12O6 contains 1.44 micro mol of glucose molecules?
I understand that to find the volume, we need to divide the moles of glucose by the molarity (mol/L) in order to get the volume. However, I am unsure of how to work with the given units and how to get the solution of 1.15 mL

MinuChoi
Posts: 100
Joined: Wed Sep 18, 2019 12:15 am

Re: Fundamental G.3 Concentration

Postby MinuChoi » Wed Oct 02, 2019 10:24 pm

You can convert micromoles to moles.
The 1.44 micromoles of glucose is equal to 1.44x10^-6 moles of glucose.

Now the units are more compatible with the molarity. Divide 1.44x10^-6 moles by the given 1.25x10^-3 mol/L measurement and the mole units will cancel out.
[1.44x10^-6 moles]/[1.25x10^-3 mol/L] = 0.00115 L, which is equal to 1.15mL.

Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

Re: Fundamental G.3 Concentration

Postby Chris Tai 1B » Thu Oct 03, 2019 9:15 am

I find it easiest to first convert the micromoles to moles (for me it's personal preference) so get 1.44x10^-6 moles of glucose.

Then you can use the equation Molarity = mols/Volume to find the volume needed, and input the values you've found. Hence,
1.44x10^-6 moles / 1.25x10^-3 mol/L = V = 0.00115 L or 1.15 mL if you want to convert to mLs.


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