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R.R. Jha
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how do I calculate nmoles from mmol/L? Molecular weight is

Resolved Question:

how do I calculate nmoles from mmol/L?
Molecular weight is 68.99527
Stock solution is 0.020 mmol/L
Volume is 1mL

What is the nmole value? and how do I work it out?
Submitted: 5 years ago.
Category: Homework
Expert:  R.R. Jha replied 5 years ago.

Number of moles
= Molarity x Volume
= (0.020 mmol/L) x (0.001 L)
= 0.00002 mmol
= 0.02 nmol

Let me know if any question.

Please click Accept
Customer: replied 5 years ago.

is that the correct calculation to answer the following?

"determine the activity of NO3 reductase in leaf tissue as nmoles NO2 produced per minute per gram of tissue"?

As I was under the impression this question was asking about mass rather than concentration?

Expert:  R.R. Jha replied 5 years ago.
I guess you wish to convert nmoles to mass.

= number of moles x Molecular weight
= (0.02*10^-9 mol) x 68.99527 g
= 1.38 nanogram
Customer: replied 5 years ago.

Last question and I will accept answer.


Is nanogram the same as nmol?


I am very confused as to what the question is asking?

I created a standard curve in mmol/L but I need to plot absorbance against nmoles not mmol/L is this a mass or a concentration?


Thanks for your time

Expert:  R.R. Jha replied 5 years ago.
nmol is a numeric unit to count number of entities, just like hundreds, thousands, or millions.
1 nmol = 6.022 x 10^14

Nanogram is unit of weight.
1 nanogram = 10^-9 grams

mmol/L is a concentration, as it defines number of entities per litre of solution.
nmoles is just number of entities (atoms, ions or molecules).

Customer: replied 5 years ago.

So how do I calculate nmol from mmol/L?


Is it the first reply you sent, or the second?

Expert:  R.R. Jha replied 5 years ago.
It was the first answer I posted above.
It's easy to see that if volume is 1 mL, number of mmol/L is same as number of nmol.
Customer: replied 5 years ago.



I've really tied myself in knots with this one and I know it should be so easy just cannot get my head round it.



Expert:  R.R. Jha replied 5 years ago.

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