The aim of question is to know about molarity of ammonia solution 25% . How can we get it?
The ammonia solution (NH3) of 25% means 25 gr NH3 in 100 ml water. The molecular weight of NH3 is 17 g (N=14 plus 3H=3). If you have 17 g NH3 in 1000 ml H2O, you have a 1M NH3. If you have 17 g NH3 in 100 ml H2O, you will have 10M NH3. Hence,17g/100ml=10M so 25g/100ml=14.7
You have a 14.7 M MH3
The concentration of ammonia (aq. sol.) should, in principle, be understood as 25 wt% (as NH3). We can retrieve densities of ammonia (aq. sol.) from (e.g.) Table 2-32, in: R.H. Perry, D.W. Green, J.O. Maloney (Eds.), "Perry's Chemical Engineers’ Handbook", 7th ed., McGraw-Hill, 1997. For 25 wt% (NH3) at 20 ºC; ρ = 0.907 g/cm3 (interpolated). The molecular mass of NH3 is 17.03 g/mol. Hence, the molarity in terms of NH3 would be: 0.25(g NH3 / g aq. sol.)·0.907(g aq. sol. / cm3)·(1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 13.3 M (as NH3).
We are not given the density of the stock solution, so we assume it has the same density as liquid water (1.00 g/mL). The molarity can be known as follows:
* Mass percent of ammonia in stock solution = 25%
* Mass concentration of ammonia in stock solution =
25g/100g × 1.00g/1mL × 1000mL/1L = 250g/L
* Molarity of ammonia in stock solution =250g/1L × 1mol/17.0g = 14.7M
This is in agreement with the answer given by Iman Tahmasbian
The ammonia solution (NH3) of 25% means 25 gr NH3 in 100 ml water. The molecular weight of NH3 is 17 g (N=14 plus 3H=3). If you have 17 g NH3 in 1000 ml H2O, you have a 1M NH3. If you have 17 g NH3 in 100 ml H2O, you will have 10M NH3. Hence,17g/100ml=10M so 25g/100ml=14.7
You have a 14.7 M MH3
I think about the molar mass of ammonia answers are not really true.
In an aqueous solution NH3 turns to NH4OH and the molar mass is 35.04gr/mol not 17 gr/mol. So 1 M means 35.04gr in a liter of water and because the solution concentration is 25%, the amount of NH4OH solution to make 1M is 140ml ,approximately.
Ms Farahani is incorrect. A 25% aqueous ammonia solution is as the name implies 25% by weight ammonia, not ammonium hydroxide. It is true that this is often called ammonium hydroxide. In fact, the solution only contains a small fraction (about 0.5%) dissociated ammonia (NH4+, OH-). The important thing is to know that this material is labeled and indeed manufactured based on weight% of ammonium (MW 17). So molarity is 14.7.
Dear Dr. Nieder,
Could you please explain your result "14.7" for molarity a 25% aqueous ammonia solution? I dare say that the correct result is the molarity "13.3", as stated by Dr. C.A. Queiroz (April 14, 2019).
Thank you so much for the explanation.
Best regards,
Karel.
I believe Dr Carlos is correct since the questioner was asking a simple question regarding a standard molarity of a chemical supplied in the form of a solution i.e. 25% ammonia solution, or say 36.9% HCl conc often found in the lab. You have to take into account the density or specific gravity and calculations must also take into account of the purity (i.e. 25%).
See
https://www.nestgrp.com/protocols/trng/molarity.shtml
13.3 M is correct. Those answering 14.7 M are ignoring the lower density of 25% aqueous ammonia which is 0.905 g/ml. The CRC handbook of Chemistry and physics (62nd edition) lists the molarity for 24 wt% and 26wt% ammonia solution as 12.827 M and 13.802 M respectively. So at 25 wt% the molarity is 13.31 M.
Check https://www.vedantu.com/question-answer/calculate-molarity-and-normality-of-25-wv-nh3-class-11-chemistry-cbse-5f682447b3d25b56229a534f
Responding to Chinaza: The original question was "How to find the molarity of solution of ammonia solution 25%?" The % units were not specified but it is logical to assume the 25% referred to wt% (wt/wt) which is equivalent to 25 g of NH3 in 100 g of solution (i.e. wt/wt basis); The 100 g of solution as mentioned above has a density of 0.905 g/ml and the concentration calculates to 13.3 M as calculated by Carlos above and listed in the CRC Handbook of Chemistry and Physics. But for Chinaza's problem, the units are different: the problem refers specifically to 25% (weight to volume) so 25 g of NH3 in 100 ml of solution results in a concentration of 14.7 M. A good reminder to explicitly state the units. (e.g. 25 wt%).
Check on this example
https://www.asdlib.org/onlineArticles/ecourseware/Analytical%20Chemistry%202.0/Text_Files_files/Chapter2.pdf
Dear Muhammad Hamza . Just use the following equation :
Molarity = (10 x 25 x density of NH3) / molecular weight of NH3.
Molarity
Moles of the soln/L,
Molarity = Mass of the solute/Volume of the soln in lit.
Hence the concentration of aqueous Ammonia For 25 % (NH3) at 20 ºC; Density = 0.907 g/cm3 (average of 24 degree and 26 degree C). As attached in Table. The molecular mass of NH3 is 17.03 g/mol. Hence, the molarity in terms of NH3 would be: 0.25(g NH3 / g aq. sol.) X 0.907(g aq. sol. / cm3)X (1000 cm3/dm3)/(17.03 g NH3/mol NH3) = 13.3 M (as NH3).
Molarity = (10 x 25 x density of NH3) / molecular weight of NH3.
Hence, molarity of 25% (w/v) NH3 solution is 14.7mol /L or 14.7 M. Normality of 25% (w/v) NH3 solution is 14.7 N. Note that normality of a solution is never less than molarity of the solution.
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