G 13

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Philip
Posts: 50
Joined: Sat Sep 07, 2019 12:16 am

G 13

Postby Philip » Thu Oct 03, 2019 11:57 pm

To prepare a fertilizer solution, a florist dilutes 1L of 2M NH4NO3 by adding 3L of Water. The florist then adds 100mL of the diluted solution to each plant. How many moles of nitrogen atoms will each plant receive?

Hannah Lee 2F
Posts: 99
Joined: Thu Jul 11, 2019 12:15 am

Re: G 13

Postby Hannah Lee 2F » Fri Oct 04, 2019 12:10 am

To prepare a fertilizer solution, a florist dilutes 1L of 2M NH4NO3 by adding 3L of Water. The florist then adds 100mL of the diluted solution to each plant. How many moles of nitrogen atoms will each plant receive?


For this equation, you would need to use M1V1 = M2V2 in order to first find the final molarity of NH4NO3. 0.20 M NH4NO3 is your initial molarity, and 1.0 L is your initial volume, while 4.0 L is your final volume, and the value you have to find is the final molarity. So: (0.20 M)(1.0 L) = M2 (4.0 L), and the final molarity of NH4NO3 turns out to be 0.050 M. Since the problem is asking for moles, you multiply it by volume to convert to moles: 0.050 mol/L (0.100 L) = 0.0050 mol NH4NO3.

Since each NH4NO3 atom contains 2 N atoms, you would use mole ratios to determine the moles of nitrogen atoms: 0.0050 mol NH4NO3 (2 mol N / 1 mol NH4NO3) = each plant receives 1.0 x 10-2 mol N atoms.

AnvitaaAnandkumar_1B
Posts: 56
Joined: Fri Aug 02, 2019 12:15 am

Re: G 13

Postby AnvitaaAnandkumar_1B » Fri Oct 04, 2019 12:12 am

To solve this we use the dilution law : M1V1 = M2V2.
M1 = 0.20M . M2 =
V1 = 1 l . V2 = 1 + 3 = 4 l
Substituting in the equation, we get M2 = 0.05M
Molarity = (No. of moles) divided by (volume in litres)
Therefore,
0.05 = (n x 1000)/ 100 = 0.005 moles of NH4NO3
Each mole of NH4NO3 has 2 moles of N
therefore, no. of moles of N = 2(0.005) = 0.010 moles of N

Sean Sugai 4E
Posts: 56
Joined: Wed Sep 11, 2019 12:17 am

Re: G 13

Postby Sean Sugai 4E » Fri Oct 04, 2019 1:30 am

To solve this equation, use the dilution equation, M1V1=M2V2, in which M1 and V1 represent the initial molarity and the volume of the solution, respectively, and M2 and V2 represent the final molarity and volume, respectively.

The problem provides M1 (0.2M NH4NO3), V1 (1L), and V2 (4L; because the florist is adding 3 L of water into the 1 L, the final volume then becomes 4 L). Plugging the given values into the dilution equation, the equation looks like: (0.2M NH4NO3)(1 L)=M2(4L). By dividing the left side of the equation by 4 L, M2 becomes isolated, thereby allowing us to solve for the final concentration of the solution, which turns out to be 0.05M NH4NO3.

Now we know the final concentration of the solution. Looking back at the problem, the question isn't asking for the final concentration of the solution, but rather for the number of moles of nitrogen atoms each plant will receive, which means that we have to find out the number of moles given the molarity and new volume. Recall that the florist wants to give 100 mL of the diluted solution to each plant, which means that 100 mL (0.1 L) becomes our new volume.

To solve for the number of moles, use the MV=n, in which M represents the molarity, V represents the volume, and n represents the number of moles. By plugging in the molarity we just found out, 0.05M NH4NO3, and the volume in which the florist is putting the solution into, 0.1 L, the equation becomes (0.05M NH4NO3)(0.1L)=0.005 mol NH4NO3. Now, using the given moles, we have to find out the number of moles of nitrogen in the compound. To do this, we look at the compound and use dimensional analysis to cancel out units. Looking at the compound, there are 2 moles of nitrogen for 1 mole of NH4NO3.

In other words, (0.005 mol NH4NO)(2 mol N/1 mol NH4NO3)=0.010 mol N or 1.0 x 10^-2 mol N.

Therefore, 1.0 x 10^-2 mol N is given to each plant.


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