Fundamentals G13

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Palmquist_Sierra_2N
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Fundamentals G13

Postby Palmquist_Sierra_2N » Wed Sep 28, 2016 5:26 pm

I'm really struggling with understanding molarity problems. Could someone explain how do G.13. To prepare a fertilizer solution, a florist dilutes 1.0 L of 0.20 M NH4NO3 (aq) by adding 3.0 L of water. The florist then adds 100 mL of the dilutes solution to each plant. How many moles of nitrogen atoms will each plant recieve?

Vivian Wang 3J
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Re: Fundamentals G13

Postby Vivian Wang 3J » Wed Sep 28, 2016 5:51 pm

In this problem, you are given initial volume (1.0 L) and initial molarity (0.20 M NH4NO3). When the florist adds 3.0 L of water, the solution is diluted, and therefore will have a new volume and new molarity. The new, or final, volume of this solution will be initial + addition water, which is:
1.0L + 3.0L = 4.0 L
However, in order to find moles of nitrogen atoms, we will need to find the molarity (moles/Liter) of the new solution. To calculate new/final molarity, you can use this equation: M(initial)*V(inital) = M(final)*V(final), where M is molarity and V is volume.
Plug in initial volume (1.0 L), initial molarity (0.20 M), and final volume (4.0 L) to get: (1.0L)(0.20M) = (4.0L)*V(final)
Cross-multiply and solve for final molarity, which equals: 0.050 M NH4NO3.

This means for every liter of water, there are 0.050 moles of NH4NO3 present, but the florist only uses 100 mL per plant.
Calculate for moles of NH4NO3 using this formula: (number of moles)/(volume in liters) = molarity. (100 mL = 0.1 L)

So: (number of moles)/(0.1 L) = 0.05 M
and number of moles of NH4NO3 = 0.0050
Last edited by Vivian Wang 3J on Thu Sep 29, 2016 7:00 am, edited 1 time in total.

AngelaWang_1F
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Re: Fundamentals G13

Postby AngelaWang_1F » Wed Sep 28, 2016 8:42 pm

The question is asking for the moles of Nitrogen atoms however, so there are 0.05 moles NH4NO3 for 100mL of the diluted solution, but since each NH4NO3 contains 2 N atoms, there would be 0.01 N atoms.

Aprice_1J
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Re: Fundamentals G13

Postby Aprice_1J » Tue Oct 08, 2019 11:15 pm

When I solved for the moles of the diluted solution I got .005 (.05M/.1L) but I am confused as to why you multiply it by two. I know that it is the moles for NH4NO3 and that there are two nitrogen atoms but what is the reasoning for multiplying the moles by two. Is it that in the ratio there are 2N for every 1 NH4NO3? I feel like that doesn’t work because the N are coming from NH4O3.

Rida Ismail 2E
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Re: Fundamentals G13

Postby Rida Ismail 2E » Wed Oct 09, 2019 3:42 pm

Aprice_1J wrote:When I solved for the moles of the diluted solution I got .005 (.05M/.1L) but I am confused as to why you multiply it by two. I know that it is the moles for NH4NO3 and that there are two nitrogen atoms but what is the reasoning for multiplying the moles by two. Is it that in the ratio there are 2N for every 1 NH4NO3? I feel like that doesn’t work because the N are coming from NH4O3.


Every NH4NO3 has 2 nitrogen atoms, therefore you must multiply by 2 to get the moles of just nitrogen in this problem

Hannah_1G
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Re: Fundamentals G13

Postby Hannah_1G » Wed Oct 09, 2019 9:14 pm

I am having trouble with this concept as well, the amount of individual moles in a molecule has the same ratio as moles of the molecule? So 2 nitrogen moles are always in 1 mole of NH4NO3?

Camille 4I
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Re: Fundamentals G13

Postby Camille 4I » Thu Oct 10, 2019 12:07 am

Is there a way to count for two nitrogen atoms in one mole of NH4NO3 before the final calculation?


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