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### Homework Problem in Section G

Posted: **Mon Oct 08, 2018 5:32 pm**

by **Keshav Bhatnagar 1H**

Can someone please explain G13?

"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."

### Re: Homework Problem in Section G

Posted: **Mon Oct 08, 2018 6:02 pm**

by **Samantha Man 1L**

So basically you have to use the M1V1=M2V2 equation to solve this by substituting you initial molar concentration and volume of the solution prior to dilution into the left side of the equation and then substituting your final volume (4l) into the right side so that you can solve for your final molarity (M2). Once you have that value, that represents the molar concentration of the diluted solution. So you use that concentration and multiply it by the 100 ml to find how many moles of nitrogen atoms are added to each plant.

(1l)(0.2M)=(M2)(4l)

M2=

M2*(100ml)= answer

### Re: Homework Problem in Section G

Posted: **Mon Oct 08, 2018 6:05 pm**

by **Samantha Man 1L**

Oh shoot also forgot to add that you have to multiply your final answer by 2 since there are 2 N atoms in NH4NO3. my bad.

### Re: Homework Problem in Section G

Posted: **Tue Oct 09, 2018 9:10 pm**

by **Sophia Fox 4B**

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.