Help with G13 Homework Problem

Moderators: Chem_Mod, Chem_Admin

Varsha Sivaganesh 1A
Posts: 51
Joined: Thu Jul 13, 2017 3:00 am

Help with G13 Homework Problem

Postby Varsha Sivaganesh 1A » Mon Oct 02, 2017 7:57 pm

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

So far, I did M(initial) x V(initial) = M(final) x V(final). For the values I plugged in 0.20M, 1.0L, and 4.0L to solve for the M(final) value. I ended up with 0.050M, which is correct according to the solutions manual.

But I am getting confused on what to do afterwards. The solutions book says that "Moles of NH4NO3 each plant receive = 5.0 x 10^-3 mol. How did they get this answer?

Ammar Amjad 1L
Posts: 54
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time

Re: Help with G13 Homework Problem

Postby Ammar Amjad 1L » Mon Oct 02, 2017 8:07 pm

Hey,
So you found the number of moles of NH4NO3, but the question is asking how many moles of nitrogen will each plant receive. To get the amount of moles of nitrogen all you would have to do is multiply 5 x 10^-3 mol of NH4NO3 x 2 mol N/1 mol NH4NO3, which gives you the answer 0.01 mol of N. (There are two Nitrogens in the compound, which is why you multiply by 2 ). I hope this helps.
Last edited by Ammar Amjad 1L on Mon Oct 02, 2017 9:01 pm, edited 3 times in total.

K Stefanescu 2I
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am

Re: Help with G13 Homework Problem

Postby K Stefanescu 2I » Mon Oct 02, 2017 8:08 pm

Your solution is correct in terms of molarity (M), which is 0.050M NH4NO3. However, the question requires that you calculate the amount of N atoms received in moles.

To do this, start by determining the moles of NH4NO3 received by each plant. As n=MV, plug in your final values for these (I prefer to always keep my volumetric units in L so I don't get lost): n=(0.050M)(100)(10^-3 L)= 5.0 * 10^-3 moles of NH4NO3.

As the question only wants to know the moles of N atoms alone, you must use the ratio of N atoms to the compound itself as follows: (5.0 * 10^-3 MNH4NO3)*(2 moles N/1 mole NH4NO3)= 10*10^-3 mole N, or as the book states it in scientific notation, 1.0*10^-2 mole N.


Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 2 guests