Question regarding concentration units

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Homework Help Overview

The discussion revolves around determining the correct unit of concentration for a substance added to a sample, specifically focusing on the calculations involving density and concentration units. The subject area includes chemistry, particularly concentration calculations and unit conversions.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculations for determining the concentration of a substance based on its volume, density, and percentage composition. Questions are raised about the implications of different density values on the final concentration unit, specifically whether it should be expressed in mg/L or µg/L.

Discussion Status

The discussion is ongoing, with participants providing calculations and clarifications regarding the density values and their impact on concentration. There is an exploration of different interpretations of density measurements, particularly in relation to international standards.

Contextual Notes

Participants note potential confusion arising from the representation of density values in different formats, which may affect calculations. The original poster's calculations and assumptions are being critically examined, particularly regarding the density of the substance in question.

Lassy
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Hi all,

I've done some experiments and I'm trying to work out the correct unit of my final concentration. The experiment goes thus far:
I added 800 uL of a substance into a litre of my sample (800uL/L).
This substance has a density of 1020 kg/m3 and is 60% w/w.
I'm trying to work out the amount of substance per litre of sample.
After my calculations I have arrived at a number of 490 but I will appreciate some clarification on this as well as if the correct final unit of concentration will be ug/L or mg/L.

Thanks in advance.
 
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800 ##\mu L## = 0.8 cc

1020 kg/m3 = 1.02 gm/cc

(0.8)(0.6)(1.02) = 0.49 gm = 490 mg
 
Chestermiller said:
800 ##\mu L## = 0.8 cc

1020 kg/m3 = 1.02 gm/cc

(0.8)(0.6)(1.02) = 0.49 gm = 490 mg

Thanks a lot for your reply, truly appreciated.

I have one more question, what if the density was 1.020 kg/m3?

Will the final concentration still be in mg/L or ug/L?
 
I moved the thread to our homework section.

As long as you keep your units consistent, that problem never comes up.

1.020 kg / m3 * 1m3/(1000 L) * 800 µL * 0.6 = ..
 
Lassy said:
Thanks a lot for your reply, truly appreciated.

I have one more question, what if the density was 1.020 kg/m3?

Will the final concentration still be in mg/L or ug/L?
That's a pretty low density
 
Chestermiller said:
That's a pretty low density

Yes, but that's what is quoted on the data sheet for the substance.

This is the calculation I did:

Mass = D x V = 1.02mg / mL x 0.8mL / 1000 mL = 0.000816 mg/mL = 0.816 mg/L = 816 ug/L
816 ug/L x 60% = 490 ug/L
 
Lassy said:
Yes, but that's what is quoted on the data sheet for the substance.

This is the calculation I did:

Mass = D x V = 1.02mg / mL x 0.8mL / 1000 mL = 0.000816 mg/mL = 0.816 mg/L = 816 ug/L
816 ug/L x 60% = 490 ug/L
So, where did the 1020 kg/m3 come from? This is 1.02 gm/cc.
 
1.020 kg/m3 is a density lower than air. There is no liquid with that density.

Could the source be from some foreign country? Compared to English, some other languages have a swapped use of "," and "." for numbers (e.g. 12345.6 can be written as 12.345,6), so 1.020 could mean 1020.
 

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