Fundamental G.13

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Amy Luu 2G
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Joined: Wed Sep 18, 2019 12:19 am

Fundamental G.13

Postby Amy Luu 2G » Thu Oct 03, 2019 12:15 am

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 diluted solution to each plant. How many moles of nitrogen atoms will each plant receive? Solve this exercise without using a calculator.
To start I tried to find the amount of moles of NH4NO3 by multiplying 0.20 mol/L by 4.0 L ( the total volume) but am unsure why this is an incorrect step. I am very confused on how to approach this problem.

SnehinRajkumar1L
Posts: 101
Joined: Thu Jul 11, 2019 12:15 am

Re: Fundamental G.13

Postby SnehinRajkumar1L » Thu Oct 03, 2019 12:23 am

First, you must solve for the new concentration after the florist dilutes it with 3 more liters of water. Using this molarity, you can calculate the number of moles each plant receives in the 100 mL. Now, for every mole of fertilizer, there are 2 moles of nitrogen. So, using this ratio, you can solve for the number of moles each plant receives.

Andrew Jang 4D
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Joined: Sat Aug 17, 2019 12:18 am

Re: Fundamental G.13

Postby Andrew Jang 4D » Thu Oct 03, 2019 12:25 am

The florist dilutes the 1.0 L of 0.20 M NH4NO3 with water, so the amount of moles of NH4NO3 is unchanged; the molarity changes from 0.20 M to 0.05 M because the volume goes from 1.0 L to 4.0 L. Each plant receives 100. mL, or 0.1 L, of solution, meaning they receive 0.005 mol of NH4NO3. In each mole of NH4NO3 there are 2 moles of N, so there would be 0.005*2 or 0.01 moles of N given to each plant.

Matthew Chan 1B
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Joined: Sat Sep 07, 2019 12:16 am

Re: Fundamental G.13

Postby Matthew Chan 1B » Thu Oct 03, 2019 12:28 am

I believe the problem says 0.20M, not 0.20 moles. 0.20 M is the molarity of the solution, aka concentration. Molarity is mol*L^-1.

We would actually need to set up an M1V1 = M2V2 equation. We know that we have to determine the M2 value, since we are given the starting concentration (0.20M) as well as the V1 value, which is 1.0 L. We also know the V2 value, which is calculated by adding 3.0 L and 1.0 L, to give us 4.0 L water. This is the final volume, since we added 3.0 additional liters of water to the 1.0 liter that we originally started with. We then solve for M2, which should give us 0.05 M.

We then take 0.05 M and multiply it by 0.100 L, since each plant receives the 100. mL of the diluted solution. This gives us 0.005 mol of NH4NO3. We then take the 0.005 mol of NH4NO3 and multiply is by 2, since there are two moles of nitrogen in one mole of NH4NO3. This gives us 0.01 moles of N. Hopefully this makes sense!

Camellia Liu 1J
Posts: 51
Joined: Sat Aug 24, 2019 12:15 am

Re: Fundamental G.13

Postby Camellia Liu 1J » Thu Oct 03, 2019 12:30 am

You would actually divide 0.2 by 4L, since the M1V1=M2V2 relationship would leave you with (1)(0.2) = (4)(M2). By doing that, you'd get M2 = 0.05 M NH4NO3.

Then you can calculate the number of moles of NH4NO3 by solving 0.05 = (# of mols) / (0.1 L), which would give you 0.005 mols of NH4NO3.

So the final step would just be to multiply (0.005 mol NH4NO3)(2 mol N/1 mol NH4NO3), since each NH4NO3 contains 2 nitrogen atoms, to get your final answer.

[Edit: Sorry, by the time I figured out how to post this there were already other explanations.]


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