G13 Homework Problem

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JasonNovik3A
Posts: 21
Joined: Fri Sep 29, 2017 7:07 am

G13 Homework Problem

Postby JasonNovik3A » Thu Oct 05, 2017 9:54 am

This is the problem. To prepare a fertilizer solution, a fl orist dilutes 1.0 L of 0.20 m NH4NO3(aq) by adding 3.0 L of water. The fl orist 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.

I looked at the answer book and I'm completely confused. Can someone explain this problem to me please? How are you supposed to divide it amongst the plants if they don't even give you the number of plants?

ConnorThomas2E
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am

Re: G13 Homework Problem

Postby ConnorThomas2E » Thu Oct 05, 2017 10:04 am

For this problem you first determine the concentration of the NH4NO3 with it's new volume:
.20M(1.0L) = .20mol
.20mol/(3.0L+1.0L) = .05M

After finding the concentration of the solution, it doesn't matter how many plants there are. It is asking how many moles one of the plants receives, so you just need to determine how many moles are in 100ml of the NH4NO3 solution:
.05M(.100L) = .005mol

Lastly, since the problem is asking for the moles of nitrogen, not the moles of NH4NO3, you need to calculate the moles of nitrogen in the solution
.005mol(2mol N/1mol NH4NO3) = .010 mol N

Alex Kashou
Posts: 38
Joined: Fri Sep 29, 2017 7:07 am

Re: G13 Homework Problem

Postby Alex Kashou » Thu Oct 05, 2017 10:08 am

Ok so to start this problem, you need to see that the solution was .2M of just 1L of solution. However, the florist added 3 more L to the solution. Thus, there are 4L total in the final solution. Using M1V1=M2V2, you can plug in the variables and get your answer. You should get .05M for the 4L solution. Then, he adds 100mL of solution to each plant. Because we know M=moles/liters, we can multiple .05M*.100L because the units will cancel ((mol/liter)*(liter)=moles). Then you will have how many moles of NH4NO3. In this molecular formula, there are 2 moles of N for every one mole of NH3NO3. Thus, using stoichiometric variables, you can get mol N.


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