Considering any volume 'v', acetone as liquid and required concentration 'C'. What is the formula to calculate required amount of liquid needed to maintain the concentration 'C'?
When you have a very diluted solution does not matter if you measure ppm as mg per liter or as mg per kilogram. This way 10ppm means, for exemple, 10mg per liter. Considering that 215 cm3 iquals 0.215 liter you have 10 mg / liter times 0.215 liter, iquals 2.15 mg of acetone.
Thank you very much sir #SILAS DERENZO
What about the atmospheric pressure,temperature, liquid density and gas density of acetone. Is it required to consider them also?
Sorry I just notice your problem: it is a gas phase. OK. Pressure and temperature are important. For low pressure (less than 5 atm), assuming ideal gas phase,
P. V= n.R.T
with p expressed in atm, V in cm3 and T in K, R = 1 (atm)x22400(cm3)/273K/1(gmol)=82,05 atm cm3/K/gmol.
If you have 215 cm3, and supposing 25C (or 293K) and 1 atm, the total moles of the misture is
n= 1x215/82,05/293=0,0089 moles of the mixture
In this case, ppm means volume basis, or the same in mol basis. 10 ppm equals 10/1,000,000= 0,00001x 0,0089= 0,000000089 moles of acetone.
multiplying by molar mass, (58g /mol) you need to inject 58x 0,000000089=0,000005162 g, or 0,005162mg of acetone.
It is really too low mass!
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