G.13

Moderators: Chem_Mod, Chem_Admin

Farah Abumeri 3F
Posts: 95
Joined: Wed Sep 30, 2020 9:51 pm
Been upvoted: 2 times

G.13

Postby Farah Abumeri 3F » Fri Oct 09, 2020 2:48 pm

G.13 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.

I understand this problem until the florsit adds 100mL to each plant, then I become a little confused. Can someone explain the reasoning especially after the florist adds 100 mL to each plant? Thank you!

Lucy Wang 2J
Posts: 86
Joined: Wed Sep 30, 2020 10:09 pm
Been upvoted: 2 times

Re: G.13

Postby Lucy Wang 2J » Fri Oct 09, 2020 2:59 pm

So in the first solution, there are 0.20 moles of NH4NO3. When the florist dilutes it by adding 3.0L of water, the molarity of the solution is .20 mol/ 4 L = .05 Molarity. From there, the florist pours 100ml, or 0.1 L, of this diluted solution into each plant. So the number of moles of NH4NO3 being added is .05M * 0.1L= .005 mol NH4OH3. But since the problem is asking for Nitrogen atoms specifically, you must multiply the moles of NH4OH3 by the ratio of Nitrogen atoms within the larger molecule, which is 2:1. So, .005 mol * 2 = .01 mole of Nitrogen atoms for each plant. The final answer is then 1.00 * 10^(-2) mol Nitrogen)

Padra_Chang_1G
Posts: 8
Joined: Wed Sep 30, 2020 9:57 pm

Re: G.13

Postby Padra_Chang_1G » Fri Oct 09, 2020 6:39 pm

Formulas
n1=n2
M1V1=M2V2
Given:
M1= 0.20 mol/L
V1= 1.0L
M2=?
V2=4L
1.) Finding the molarity of the diluted solution
M1V1=M2V2 --->M2=(M1V1)/V2 ---> M2=(0.20*1)/4 ---> M2= 0.05 mol/L
2.)Each plant receives 100mL of the solution
100mL=0.1L
We need to find the moles of NH4NO3:
n=MV ---> n=0.05 mol/L *0.1L ---> 0.005molNH4NO3
3.) The question asks for nitrogen atoms so using molar ratios we can get the final answer
0.005molNH2NO3*(2molN/1molNH4NO3)=0.01molN
The final answer is 0.01molN or 1*10^(-2) molN

bgiorgi_3A
Posts: 56
Joined: Wed Sep 30, 2020 9:50 pm

Re: G.13

Postby bgiorgi_3A » Fri Oct 09, 2020 7:04 pm

Hi Farah! I'm sure you probably understand how to do the calculations by now, but I just wanted to recommend drawing a diagram. When I got stuck on this problem, I drew a tiny beaker and buret to help me visualize what I was actually doing. I recommend drawing out concepts if you are ever completely lost!


Return to “Molarity, Solutions, Dilutions”

Who is online

Users browsing this forum: No registered users and 1 guest