Hi Kumar,

I suggest you should prepare 1.2 L of approximately 1 vpm (volume per million) as usual in gas analysis. Your liquid ammonia solution is not the best choice since it is not sufficiently accurate with the (w/w) statement. I suggest taking ammonium chloride and preparing the ammonia (NH3) with about 2 drops of a 10 % (w/w) KOH solution. If NH3 gas is treated like an ideal gas (= 22,4 L per mole) you can proceed as follows: weight 28,66 mg of NH4Cl in a 1 L volumetric flask. From this solution you transfer 0.1 ml into your gas chamber and add the 2-4 drops of the KOH solution and close your chamber, wait for distribution. The KOH will free NH3 from the ammonium chloride and prevents any dissolved NH3 in the liquid. Anyway, you will not get an exact 1 vpm gas, due to the ideal gas assumption and adsorption phenomena at the chamber walls. The ultimate way would be to prepare more than 10 L and take several liters (depending on the problem of acid-bas titrations with very dilute standard solutions – below 0.0001 M/L the endpoint will be difficult to detect) in order to purge them through a known amount of very dilute H2SO4 standard solution and make a back titration with NaOH or KOH and determine the exact amount of NH3 transformed into ammonium sulfate (usual acid base titration).

Hi Kumar,

I suggest you should prepare 1.2 L of approximately 1 vpm (volume per million) as usual in gas analysis. Your liquid ammonia solution is not the best choice since it is not sufficiently accurate with the (w/w) statement. I suggest taking ammonium chloride and preparing the ammonia (NH3) with about 2 drops of a 10 % (w/w) KOH solution. If NH3 gas is treated like an ideal gas (= 22,4 L per mole) you can proceed as follows: weight 28,66 mg of NH4Cl in a 1 L volumetric flask. From this solution you transfer 0.1 ml into your gas chamber and add the 2-4 drops of the KOH solution and close your chamber, wait for distribution. The KOH will free NH3 from the ammonium chloride and prevents any dissolved NH3 in the liquid. Anyway, you will not get an exact 1 vpm gas, due to the ideal gas assumption and adsorption phenomena at the chamber walls. The ultimate way would be to prepare more than 10 L and take several liters (depending on the problem of acid-bas titrations with very dilute standard solutions – below 0.0001 M/L the endpoint will be difficult to detect) in order to purge them through a known amount of very dilute H2SO4 standard solution and make a back titration with NaOH or KOH and determine the exact amount of NH3 transformed into ammonium sulfate (usual acid base titration).

ppm = Molarity of ammonia X molecular weight of ammonia X 1000

ppm is equivalent to mg/L.. is it possible for you to convert the unit to this?

Hi Kumar,

if it's possible for your research group, I suggest to buy anidrous solid ammonia, with which you can easly prepare 1 ppm. In fact, liquid ammonia 30% (w/w) has not an exact concentration, so it should be very difficult to prepare an exact standard. Alternatively, also ammonium hydroxide solutions with a cerified molarity are a good (and generally cheap) choice...Anyway, in these cases, I always suggest to titrate the final solution.

Hello Kumar,

I am assuming that by 1 ppm you mean 1 ppm v/v gaseous and not 1 mg/L, which is not exactly the same.

As has been pointed out by Karl Camman, there are better methods than starting with a 30% aqueous ammonia solution. For accuracy, starting with ammonium chloride and preparing a standard aqueous stock solution of the chloride is preferred. The major problem is, as has been noted, adsorption of gaseous ammonia onto the inner walls of your chamber, which will reduce the gaseous concentration. You cannot generate a "primary standard" concentration in this way.

A better approach would be to continuously meter a higher, known concentration of ammonia gas, generated as described, into a nitrogen stream passed through the chamber at a dilution rate sufficient to produce the final 1 ppm concentration desired. Allow the system to operate for a short period to achieve adsorption steady state. The concentration inside the chamber will approach a steady state concentration close to that desired. This of course assumes that your chamber and gas feed system is designed to promote turbulent mixing of the gas mixture within the chamber. I have had good success with such dynamic systems, but have rarely had success with static dilution systems because of these adsorption effects.

I agree with Charles: dynamic mixing and diluting is the more accurate method. The ammonia stock gas with a higher concentration can be analyzed much easier and with higher accuracy. The precision of the final gas will then be limited by the accuracy of the two gas flow meters of the stock gas and the nitrogen gas if the mixing is perfect.

If the final ammonia concentration must not be exactly 1 vpm an alternative method could be also to place an ammonia source (e.g. your concentrated NH4OH solution) in a mermeation tube ( http://www.vici.com/calib/perm_dyna.php). in the nitrogen stream. At constant pressure and temperature the leakage rate of ammonia is constant and you have plenty of time to collect the resulting gas in larger amounts to determine its ammonia content.

30% Ammonia solution w / w contains 30 g / g or 30 g / ml (30 kg / l), so the rule of three for 1.2 liter leads to: 30kg x 1L / 1.2 L = 25kg or 25g/ml or 25g / g or 25 ppm. That said, it seems a lot and I do not see the advantage of using ppm (concentration unit reserved for traces).

Hi, Kumar

I agree with Karl Cammann

Our isntitute has a very long and large expertise with measuring ammonia

see for instance

Wyers, G.P., Otjes, R.P., Slanina, J., 1993. A continuous-flow denuder for the

measurement of ambient concentrations and surface-exchange fluxes of

ammonia. Atmospheric Environment 27, 2085–2090.

A major problem with gaseous ammonia in chambers is that it sticks to the walls of the chamber and to the tubing in whihc it is brought in and pumped out for measuring.

So it would be good that you tell us for what purpose you use the ammonia.

Also: what precision do you need in your tests and how are you measuring the gaseous ammonia

Streight and easy answer to your question is 30% ammonia is equal to 300,000 ppm. To get 1 ppm in a 1.2 L Chamber. Set up the equation C1V1 = C2V2 where C1 = 1 ppm, V1 = 1.2 L, C2 = 300,000 ppm. Therefore V2 = C1V1/C2 = 4E-6 L or 4E-3 mL = 4 uL. Four uL can be measured fairly accurately with a 5 uL syringe. This is a nominal value, which may be accurate enough. I am not sure exactly how to go about actually measuring ammonia in air with some sort of instrumentation.

Thomas

The least expensive way is using a denuder tube thru which you pump the air of the chamber. Then rinse the adsorbed ammonia off and measure with a wet ammonium analysis system.

The inner wall of the deunder tube must be coated with for instance oxalic or phosporic acid to capture the ammonia.

http://pollutantdeposition.defra.gov.uk/ammonia_methodology

http://www.sciencedirect.com/science/article/pii/1352231094900078

SEmi-continuous monitors based on denuders: AIRMONIA or AMANDA commercial detectors

More expensive methods are based on optical techniques, like DOAS or photoaccoustics.

A 30% (W/W) solution means that you have 30 g of ammonia dissolved in 100 g of water. It means that you have 300 g of ammonia per liter of water or 300 000 mg of ammonia per liter. A solution of ammonia with a concentration of 1 ppm means that you have dissolved 1 mg of ammonia in 1000 mL of water.

Then you have that V2=(c1 x v1) / c2

V2 = (1 mg/L x 1000 mL) / 300 000 mg = 1/300 mL or 3.333 micro liter.

Alain

This question is NOT about ppm of ammonia in water but ppm in air.

As mentiioned above in one of the answers, this measn volume by volume ppmv

For ammonia it is slightly less than 1mg of ammonia in his chamber of 1,2 m3.

Correction: I just noticed that the flask volume is only 1.2 liters. This would mean 1 microgram of ammonia.

For me it is mysterious what you want to do with this gaseous ammonia in such a small flask.

I ask you, Gaustav Kumar, this because I worked with an air volume of 30 cubic meter and concentrations of 1 ppb

Good discussion!

Let us keep our units clear to get 1 ppm ammonia in in 1.2 L, you will need about 1.2 mg of ammonia for a weight to volume. But you are correct Harry, the correct measure should probably done on a v/v basis and not a w/v basis. In order to get this right you would have to know the contributing vapor pressure of 1.2 mL of ammonia in 1.2 L of air assuming that we are not dealing with a vacuum here.

I agree with the collegues, but what do you really want to do with this calculation?

In the above answers there is a confusion between ppm in water versus ppm in the gaseous phase. Parts per million (ppm) in the water phase is the same as mg/L (this latter designation is preferred). Parts per million in the gaseous phase (proper abbreviation ppmv) is quite different. It is actually a mole fraction on a per million basis. 1 ppmv means 1 molecule of the specified substance per million molecules of gas.

ppm, is part per million, is is in ug/mL, or ng/uL. then you can calculate

Gaurav Kumar

Some calculation:

in botlle V=1.2L 1 ppmv NH3 is V_NH3 = 1,2*10^-6 L i.e. 0.000 0012 L = 1.210^-9 m3

Ideal gas law:

p*V_NH3=((m_NH3)/(M_NH3))*R*T where: p - presure [Pa], V - volume [m3], m_NH3 - mass of ammonia [kg], M_NH3 - molecular weight of NH3 []kg/kmol], R - universal gas constant [= 8314 J/kmol*K], T - temperature [K]

therefore:

m_NH3 = (p*V_NH3*M_NH3)/(R*T)

on the other side:

c_%w = m_NH3/m_solutionNH3

so

m_NH3 = m_solutionNH3*c_%w

then

m_solutionNH3*c_%w = (p*V_NH3*M_NH3)/(R*T)

then

m_solutionNH3 = (p*V_NH3*M_NH3)/(R*T*c_%w)

if p = 101325 Pa (standard pressure), V_NH3 = 1.210^-9 m3, M_NH3 = 17.09 kg/kmol, R = 8314 (J/kmol*K), T = 290 K (~17 C or ~62 F) and c_%w = 30% i.e.0.30

then

m_solutionNH3 = 2.87*10^-9 kg (!)

You have a problem - is strong difficulte weigh above 3 microgramms.

Solution: Try dilute 30% solution

I like to refer to the website of the Environmental Protection Agency (of the US).

The concentraions of traces of air pollutants is tabulated in ppm and ppb.

This is on a volume by volume basis. BTW, this is the same as moe to mole because all gases have the same volume per mole (gas law).

I thus inferred that the question was according to the international convention for trace gases in air. see table for CO/NO2 etc:

http://www.epa.gov/air/criteria.html

And here the health issues of ammonia in air IN the unit ppm!!

• Animal Health Issues

• Similar to human effects (eyes, nose, throat irritation/damage)

• Swine:

• reduced weight gain of 12-32% at NH3 levels of 50-100 ppm

• increase stress, lowered resistance to infectious disease

• recommend levels below 20 ppm (others suggest 10-15 ppm)

• Cattle:

• pulmonary effects most prominent (reduced lung function)

• decreased mucociliary transport

• Poultry:

• reduced body weights at 25 ppm

• respiratory irritation, predisposition to infectious disease, and

cornea/conjunctiva inflammation (keratoconjunctivis) at 50 ppm

GOOGLE 1st hit:

epa ppm nh3 air pollutants

Before you invent the wheel (again)

Here is a site for the conversion of ppm into milligram per m3 of air and vice versa

http://www.skcinc.com/converter/converter.asp

ppm= ( gram of solute/ gram of solution) X 10 to the power 6. Examples: Liquids: 1mg of salt in 1 liter of solution is 1 ppm concentration solution. Solids: microgram/gram is 1ppm.

Bommanna

Please look at the question again:

he asks: literally "to make 1 ppm after evaporation"

so it is NOT a liquid solution, bur vapour in air

Harry Thank you for the clarification. You gave an excellent resource as an answer for the question.

ppm means mg/L. For 1.2 L, the liq. ammonia needed is 1.2 mg in order to make concentration as 1 ppm. The conc. of liq. ammonia is 30% w/w. Hence 1.2/30% i. e. 4 mg of ammonia is required to achieve the desired concentration.

mr (Balkrishna Ginde) Dr

You obviously did not look further is this toipic, see the replies by dr Loganathan!!

I appreciate with Balkrishna Ginde Dr. in term of calculation. However more precautions should be taken during preparation otherwise you cannot get as accurate as required

mr Hamidar

the querstion was about ammonia in the gaspahse not in solution; please read the question more carefully. The question is literally quote:

"to make 1 ppm after evaporation"

1 ppm means 1 mg. Of the substance in 1000 c.c. I.e. 1 liter solution

30% ammonia solutions means 30 gm. Ammonia in 100 gm . Of water or approximately 100 c.c. Solution that means 30% ammonia is = 300 ppm. Ammonia

in the question it has not been mentioned the name of the solvent .in general ammonia is sold in the market as 30%water solution . Now solve your problem

Mr Majumbur

Please read the question more carefully

THe question is about ammonia in the gas phase not in solution. The question is literally, quote:

"to make 1 ppm after EVAPORATION"

Weigh 3.3 mg of the ammonia and dissolve in 1 L of de-ionized water. If you want to prepare a larger volume use must increase the quantity of ammonia and water proportionally. E.g. if you want to prepare 2 L of 1 ppm solution then weigh 3.6 mg of ammonia and dissolve it in 2 L.

Sorry ,extremly sorry. 30% w/w ammonia in water soln means that the soln. contains 30 gm. of ammonia in 100 gm of water. which is approximately 100 cc. of water that means 30 x 1000 mg. of ammonia in 100 cc . of water.i.e. 30 x 10000 mg. of ammonia in 1000 c of water. that the con. of the soln. is 30,0000 ppm of ammonia

Dear Mr. Majumdar,

You are right. The comment of Mr. Brink, whether in liquid phase or in gaseous phase has no relevance here. The calculations given by me are correct. Hwever as rightly stated by Mr. Hamidar, precautions need to be taken for accurate weighment of ammonia liq.

PPM (Parts per million) is a measurement used today by many customers to measure quality performance.

Definition: One PPM means one (defect or event) in a million or 1/1,000,000

Parts per Million by Volume (or mole) in Air

In air pollution literature ppm applied to a gas, always means parts per million by volume or by mole. These are identical for an ideal gas, and practically identical for most gases of air pollution interest at 1 atm. Another way of expressing this value is ppmv. [1]

One part per million (by volume) is equal to a volume of a given gas mixed in a million volumes of air:

A micro liter volume of gas in one liter of air would therefore be equal to 1 ppm:

Today's more and more there is an interest to express gas concentrations in metric units, i.e. µg/m3. Although expressing gaseous concentrations in µg/m3 units, has the advantage of metric expression, it has the disadvantage of being greatly influenced by changes in temperature and pressure. Additionally, because of difference in molecular weight, comparisons of concentrations of different gases are difficult. [2]

To convert ppmv to a metric expression like µg/m3, the density of the concerning gas is needed. The density of gas can be calculated by the Law of Avogadro's, which says: equal volumes of gases, at the same temperature and pressure, contain the same number of molecules. This law implies that 1 mole of gas at STP a volume of 22.71108 liters (dm3) enfolds, also mentioned as the molar volume of ideal gas. Standard Temperature and Pressure (STP) is defined as a condition of 100.00 kPa (1 bar) and 273.15 K (0°C), which is a standard of IUPAC. [3] The amount of moles of the concerning gas can be calculated with the molecular weight.

Where:

Vm =

standard molar volume of ideal gas (at 1 bar and 273.15 K) [3]

[22.71108 L/mol]

M =

molecular weight of gas

[g/mol]

For converting ppm by mole, the same equation can be used. This can be made clear by the following notation:

By checking the dimensions of the most right part of the equation, there will be found a dimensionless value, like the concentration in ppm is.

To calculate the concentration in metric dimensions, with other temperature and pressure conditions the Ideal Gas Law comes in handy. The volume (V) divided by the number of molecules (n) represents the molar volume (Vn) of the gas with a temperature (T) and pressure (P).

Where:

Vn =

specific molar volume of ideal gas (at pressure P and temperature T)

[L/mol]

V =

volume of the gas

[m3]

n =

amount of molecules

[mol]

R =

universal gas law constant [3]

[8.314510 J K-1 mol-1] or [m3 Pa K-1 mol-1]

T =

temperature

[K]

P =

pressure

[Pa]

With this equation it comes clear that the percentage notation by ppm is much more useful, because the independency of the temperature and pressure.

Parts per Million by Weight in Water

The concentration in ppm of gas in water is meanly meant by weight. To express this concentration with metric units the density of water is needed.

The density of pure water has to be by definition 1000.0000 kg/m3 at a temperature of 3.98°C and standard atmospheric pressure, till 1969. Till then this was mean definition for the kilogram. Today's the kilo is defined as being equal to the mass of the international prototype of the kilogram [4]. Water with a high purity (VSMOW) at a temperature of 4°C (IPTS-68) and standard atmospheric pressure has a density of 999.9750 kg/m3. [5]

The density of water is effected by the temperature, pressure and impurities, i.e. dissolved gasses or the salinity of the water. Even the concerning concentration of gas dissolved in the water is affecting the density of the solution. By nature there's a chance that water contains a certain concentration of Deuterium which influences the density of the water. This concentration is also called the isotopic composition [6].

Accurate calculations on these conversions are only possible when the density of the water is measured. In practice the density of water is therefore set to 1.0 ·103 kg/m3. When calculating the conversion with this value you gets:

Where:

?w =

density of water

[1.0 ·103 kg/m3]

Reference

[1] Never, N. , Air Pollution Control Engineering. McGraw-HILL, Singapore 1995.

[2] Godish, T. , Air Quality. Lewis Publishers, Michigan 1991.

[3] Cohen, E.R. and Taylor, B.N., J. Res. Nat. Bur. Stand. 92 (1987) 85-95. (International Union of Pure and Applied Chemistry (IUPAC))

[4] n/a, Kilogram. International prototype of the kilogram, www.bipm.org/en/scientific/mass/prototype.html.

[5] Marsh, K.N., Ed., Recommended Reference Materials for the Realization of Physicochemical Properties. Blackwell Scientific Publications, Oxford.

[6] www.iapws.org/faq1/isotope.htm

Read more: http://www.lenntech.com/calculators/ppm/converter-parts-per-million.htm#ixzz2GyYcrtgE

Ramesh Padodara

I like to reiterate soem of my replies. This is 1sy year air pollution course

For a gas in air the concept of ppm means parts per million expressed in volume by volume. Or moles per moles which is the same as volume by volume

Go to the site of the EPA of the United States and you see that ppm is the standard unit for gaseous air pollutants.

Remmber this topic here is Environemtnal Pollution, so the EPA reference is the one to use.

Gaurav Kumar

I HAVE ASKED THIS BEFORE WHY DO YOU WANT TO KNOW THIS.

OR IS THIS AN EASY WAY TO GET YOUR HOME-WORK DONE BY US.

I had and have the hope that Research Gate is for serious science.

Assuming that only ammonia evaporates from the aqueous solution and ideal gas law applies then the following procedure should be used to solve this problem. It should also be assumed that all ammonia in the liquid phase evaporates to the gas phase.

Cppm = (nA x 10^6)/Ntotal ==> nA = Cppm x Ntotal/10^6

Ideal gas ==> PV=Ntotal x RT ==> Ntotal = PV/RT

Hence nA = Cppm x PV/(RT x 10^6) = mA/MA ==> mA = MA x Cppm x PV/(RT x 10^6)

This is the mass of ammonia in the gas phase that would achieve the target Cppm. Now you need to calculate the mass of solution that will give you this ammonia mass mA.

By definition the percentage of ammonia in water is CAW = mA/ms ==> ms = mA/CAW

Then ms = MA x Cppm x PV/(CAW x RT x 10^6)

Where MA: molecular mass of ammonia (MA = 17.031 x 10^-3 kg/mol)

Cppm = 1 ppm

V=1.2 x 10^-3 m3

CAW = 0.3

R=8.314 J/mol.K

Assume the operation is made at room conditions then

P=1.013 x 10^5 Pa

T= 20+273.15 = 293.15 K

The above data give ms = 2.826 x 10^-9 kg

ms=2.826 ug (do you have a balance to measure this mass?).

This is clearly not practical but you may try to dilute the liquid ammonia solution first (e.g. thousand times) and/or increase the volume of the gas vessel.

Hope this helps.

Chedly Tizaoui

Chedly Tizaoui

Here we go again: this is the site that does it for you

http://www.skcinc.com/converter/converter.asp

thanks, but I prefer to use my own spreadsheet.

Hi Harry, Can we keep Air or any other vapour out because ammonia is in water solution.If it is vapour it is that of water.Next:Specific gravity of the 30% solution is0.93.It comes out that the molarity of the given solution is 16.39M.That means 16.39 moles of ammonia are there in one litre of the given solution.Which is equivalent to 379g of ammonia in 1000ml.OR 2.79mg in 1 ml.That is 1 mg of ammonia in 0.36ml of given solution.OR 1mg = 0.36ml of given solution.Now add 20% to each side to make 1.2L of solution.i.e,;1.2mg of ammonia =0.43ml of the given solution,That means 0.43ml of the given solution can make 1.2L of the 1ppm solution.QED.

A shok

I referfor the correct answer to Chedly Tizaoui

The original question is, literal quote:

"to make 1 ppm after evaporation". This means ammonia in the gas phase. The term ppm means that the ammonia is in another gas. Becasue it is in a gas it does not matter if it is air or nitrogen or what.

Please re-read the original question, at the top of this topic

Harry Sir,

Yes I have taken H2O (g) and ammonia as a substance.Is it different from air or any other gas for that matter.Presence of any other gas is not specified and we should not imagine what is not given.Water as a solvent is already given.Moreover will it make a difference if 1 ppm is in water or air or any other solvent.More so ppm depicts a ratio of particles irrespective of foreign material as long as they make a solution.

Harry, Chedly has a point .

Moreover ppm is not volume by volume.It is particle by particle consideration.However it has some equivalence through formulation.

It seems to me that the question is some what confusing. it is given that the ammonia solution is 30%w/w but the it is not stated that 30% in what solvent . If I take the case of ammonia in water then while evaporating ammonia a portion a water will be present in the vapour phase. and the amount of each constituent will depend on temp. pressure and the distribution ratio of the two. so the calculation is not so easy.. To prepare 1 ppm. of ammonia it is better to use ammonium chloride as solid.dissolved in water and can be measured accurately up to 0.50 ppm

mr Majumbar

You are right that it is not given in what solvent.

As for aamonium chloride solution: when this is evaporated, by heating fro instance both gaseous ammonia and hydrochloric acid are formed. In the chamber the gaseous acid will immediately attach to the walls and then the ammonia will react with the hydrochloric acid at the wall. This analogous to the fate of ammonium nitrate that we have studied for over 30 years and whihc we brought into chambers. The way we dis that is atomizing / nebulizing a solution and then drying the droplets. Ammonium nitrate is semi volatile and it dissociates into ammonia and nitric acid in the chamber.

to Ashok Kaushal:

for a gas particle by particle is the same as volume by volume according to the gas law. Why do you say it is not?

I have not asked to evaporate ammonium chloride but have suggested how the standard curve of determining ammonia in order of < 1 ppm can be estimated accurately

Hi Harry ,Indirectly yes.I did not say it can not be equated.I wanted to stress the point that ppm starts from particles, it can be settled there itself, no need to convert it to volume and back to particles.Harry this problem has become very interesting in the face of uncertainties of medium.Can it be logically concluded.

Ashok

please ask Gaurav Kumar directly what solvent he uses; he is very reluctant to answer my questions to him

Thank all of my Respected Sir for your valuable suggestions

This is liquor ammonia (Ammonia in water)

Hi gaurav we can calculate the concentration as follow :

ppm= mg/l wt= 0.3 mg vol.=1.2 l so ppm=0.3/1.2=0.25 ppm ammonia

Hi Harry ,

Gaurav has made it clear, it is ammonia in water.And correct answer is also already there.

Jonathon,

at least I enjoyed the twist and vagueness of a realistic situation.

Gaurav,

I hope you got the reply to your satisfaction.

Chedly may also refer to my solution after his solution.

ppm= mg/Kg or mg/L or ug/g

Sorry Teodoro

We had a very long discussion here already.

The question is to produce ammonia in the GAS-PHASE by evaporating a solution. For a gas ppm is defined on a volume by volume basis. see for instance the website of the OSHA, that provides exposure limits for the concentration of air pollutants and in this case ammonia in the unit ppm or milligram / m3

Harry

The website

http://www.osha.gov/dts/chemicalsampling/data/CH_218300.html

Gaurav 's interaction is required here.

If 1 part per million space units is ammonia what occupies the rest of the volume in a L.

We are not measuring the concentration of ammonia in air.See the original question Harry,pl.

Hi Gaurav

ppm in gas solution probably refer to micro liter in 1 lit gas solution therefore 1ppm ammonia in 1.2L volume is calculable,

you can1.2 micro litre ammonia evaporated in 1.2L chamber.

Refrence Book : QUENTITATIVE CHEMICAL ANALYSIS, BY: Daniel C. Harris

I assume that refrence solution of ammonia is 100% pure, you can calculate requier volume with density and percent of refrence solution.

this is only a suggestion

First of all, I think this discussion should be closed as it is clearly just a grad student who wants us to make his homework.

Secondly; many thing are postulated here, about what ppm is. Ppm is parts per million, the parts can be whatever you want but it should be mentioned (though Hadi’s suggestion is a bit strange, defining ppm as one volume of 30%amonium liquid solution per million gas volumes, and a solution of 100% ammonia is a contradiction in terminis ). If it is not mentioned the most logical to consider it as mole per mole, and for an ideal gas this is the same as volume to volume. The next problem is the 30% this is most probably mass to mass (but it could also be mass to volume, it should be mentioned).

So let’s say that the solution is in mass to mass and the ppm required is in volume to volume and that everything is at standard temperature and pressure. So you need to obtain 1.2 µl of NH3 gas in 1.2 L (the composition of the rest do not matter, only the volume), 1.2µl NH3 = (/22.3L/mol) 53.6nmol of NH3 = 912ng of NH3 so you need to put 3.04 µg of the 30% solution in the 1.2L jar.

Of course this is not what I would use to make a calibration gas (as indeed NH3 sticks every where etc…), but this is the right answer to the question asked.

mr Bode

I like to refer you to the answer by dr Cammann above, who gives the same calculation. The questioneer is not a student but somewaht who mentioned earlier who wants to develop a new biosensor for ammonia.

Dr Cammann also provided in this subject some weeks ago the name of a company that has a calibration gas in a cylinder.

3.22X10-6 mL is needed

ppm = mg/L

ppm concentration in air phase is different from the solution phase. So, ppm in air phase is not mg/L. The equation is as following:

mikrogram ammonia/m3 air(1.2X10-3)= (ppmX17X1000)/24

The calculations show that the answer is 3.2X10-6 mL

Once again

The question is TO MAKE 1 ppm AFTER EVAPORATION; please read the WHOLE question

If I say 1.2mg of dry pure ammonia in 1.2L of space at STP and there is nothing else except the vapours of ammonia.will it be called 1 ppm of ammonia ?

Ashok

You are of course right: 1 ppm means 1 molecule (or mole) in a million of total gas molecules (moles).

Dr. Brink I need your e-mail ID as I need to discuss few more things on biosensors, if you allow for the same....

Sorry but I am not the right perosn on biosensors; I advice you to approach dr Cammann

If you need absolutly one ppm of 33% ammonia solution you can evaporate 3.3x1o-3 ul sawsan

Sawsan

Do you know the size of what you propose: how big is one droplet from a pipette or burette

Answer: 30-50 milliliter, with that uncertainty

I think that Dr. Cammam give a good desription

1 ppm = 1 mg/L

g/mL = mg/L

1g/ 1000mL = 1000 ppm solution

ppm= 10^-6 , ppb = 10^-9

Krithika

NO,

This question is ppm in the GAS-PHASE, see the answer by dr Cammann

Hi,

I have a question regarding to liquid formaldehyde. I have 37% formaldehyde. If i want to do the chemical sensing so how i have to calculate the ppm in this case.

Thanks.

Multiply by 37/30 and then again with the ratio of the molar weights of formaldehyde and ammonia

Hi,

I have a question, how can I calculate 300 microgram of Nitrogen (N) in ammonium sulphate?

2.0 µL/L formaldehyde is how much ppm?

Formaldehyde : 812 kg/m³

30,02598 g/mol

Gaurav said liquid ammonia and not the ammonical liquid. It could be Liquefied 30% ammonia in nitrogen under pressure. Then calculation becomes easy.

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