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Topic: Homework help-concentrations, molar masses etc  (Read 1046 times)

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Offline Number_are_not_my_thing

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Homework help-concentrations, molar masses etc
« on: December 20, 2020, 01:02:32 PM »
I'm doing a module in medicinal chemistry figuring out concentrations and such is not my strong suit and I get easily stressed with it and forget how to do it :( Also I haven't done anything like this in about 6 years!

1) for this one I need to figure out what volumes of each solvent obtained in an experiment  I would need to make stock solutions of 40mg/ml. The solvent/extracts obtained are 76mg and 37mg. I'm assuming the desired volume is 1000ml.

I need to use V1= (C2 x V2)/ C1, right? But I feel like I'm over complicating it.

2) I'm pretty sure I got this one right. What volume of stock solution would I need to get a concentration of 1mg/ml in a total volume of 200 μl? I got 0.005ml/5μl

3) Erythromycin A has a molecular mass of 733.93g/mol. What mass do I need to make 1ml of stock solution with a concentration of 20mM?
What formula do I need?

Offline chenbeier

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Re: Homework help-concentrations, molar masses etc
« Reply #1 on: December 20, 2020, 02:03:00 PM »
Number 1 is not clear, so you want to mix two solution contains 76 and 37 mg/l to get 40 mg/l or do want to get it by dilution of 76 mg/ l. 37 mg/l is already to less.
In first case you can use cross rule.
Number 2 will be 0,005 ml for 200μl not 5 μl.
Number3 : Calculate the mass of  20 mM. Molar mass is given.   m = n*M
Calculate this for 1 ml.

Offline Number_are_not_my_thing

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Re: Homework help-concentrations, molar masses etc
« Reply #2 on: December 20, 2020, 02:26:33 PM »
Thank you so much! You're a life saver :D
For 1, it's a separate solution for 76mg and 36mg by dilution. The question isn't very clear, I assume this is what it wants.

I've currently done 40mg/ml= 76mg/volume so volume= 76/40 =1.9mg

for 3 I got 0.0146786 mg
0.73393 mg/mM * 20 *1 / 1000


Offline chenbeier

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Re: Homework help-concentrations, molar masses etc
« Reply #3 on: December 20, 2020, 03:59:06 PM »
No,  The start solution has 76 mg/l. The end solution should have 40 mg/l
40 mg/l is the same as 76 mg/1,9 l. So you have to add 0,9 l to get this result.

Number 3 is correct.

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