Question G13 (Sixth Edition)

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Lauren Huang 1H
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Joined: Fri Sep 28, 2018 12:16 am

Question G13 (Sixth Edition)

Postby Lauren Huang 1H » Tue Oct 09, 2018 2:21 pm

Can someone please explain how to do G13? It says: "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."

Charles Gu 1D
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Joined: Fri Sep 28, 2018 12:16 am

Re: Question G13 (Sixth Edition)

Postby Charles Gu 1D » Tue Oct 09, 2018 2:31 pm

You first have to use the M1V1= M2V2 equation in order to get M2. After calculating for M2 (Molarity of NH4NO3), multiply that by 0.1 L (100 ml, given in the problem) in order to get the moles of NH4NO3. Since there are 2 moles of N for every 1 mol of NH4NO3, multiply the mols of NH4NO3 by 2 and you get 0.01 moles of Nitrogen.

Sophia Fox 4B
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Joined: Fri Sep 28, 2018 12:27 am

Re: Question G13 (Sixth Edition)

Postby Sophia Fox 4B » Tue Oct 09, 2018 9:12 pm

Hi! I just posted this same answer on another post but here goes:
So first you have to isolate for Mf. Then you do (.2M * 1.0L)/(1+3L) to find the final molarity of the NH4NO3 solution, which is .05M. Once you have that, multiply by .1L (=100mL) to find how many moles of NH4NO3 for each plant, which is .005 mol NH4NO3. Then use mole ratios to find the moles of just N. Since each molecule of NH4NO3 has 2 Nitrogen atoms, multiply the .005 mol NH4NO3 by the ratio 2 mol N / 1 mol NH4NO3. Then you get the answer: .01 mol N atom per plant.

Dana Wilks 3I
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Joined: Fri Sep 28, 2018 12:23 am

Re: Question G13 (Sixth Edition)

Postby Dana Wilks 3I » Wed Oct 10, 2018 4:32 pm

Is it necessary to use the MiVi=MfVf equation? I set up this problem by using molarity=moles/liters, which gave me the number of moles (.2). I then recalculated the molarity using the new volume of 4 L to get a molarity of .05. Using .05 M and the given information that each plant was getting .1L I solved for the moles of N.

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