Question about G.13

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kennedy_meyer_
Posts: 3
Joined: Fri Sep 29, 2017 7:04 am

Question about G.13

Postby kennedy_meyer_ » Thu Oct 05, 2017 2:47 pm

In the problem G.13, how do you determine that the concentration of the diluted NH4NO3 solution is .050 M?

704887365
Posts: 49
Joined: Sat Jul 22, 2017 3:00 am

Re: Question about G.13

Postby 704887365 » Thu Oct 05, 2017 2:55 pm

You have to use MV=MV, using (1)(0.02)=(4)(M). This will give you the concentration as 0.05M.

Maria1E
Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

Re: Question about G.13

Postby Maria1E » Thu Oct 05, 2017 2:57 pm

You use a key concept that Dr. Lavelle outlined in his lecture on 10/2: When diluting a solution, moles of solute remain the same. Therefore, you can use the formula (M initial)(V initial) = (M final)(V final), where M is molarity and V is volume. For this problem, we are told that a florist dilutes 1 L of 0.20 M NH4NO3 by adding 3 L of water. You want to find the molarity of the diluted solution. The initial molarity is 0.20 M NH4NO3 and the initial volume is 1 L. Since you are adding 3 L of water to the original volume of 1, the final volume is 4 L. Then plug those factors into the equation I mentioned above. You will get:

(0.20 M NH4NO3)(1 L NH4NO3) = (? M)(4 L NH4NO3)
0.20 = (? M)(4)
?M = .20/4
= .050 M NH4NO3

Ashley Chipoletti 1I
Posts: 20
Joined: Fri Sep 29, 2017 7:04 am

Re: Question about G.13

Postby Ashley Chipoletti 1I » Thu Oct 05, 2017 3:00 pm

In this problem, you would use M(initial)V(initial)= M(final) V(final) to find the concentration. For M(initial), plug in 1.0 L and for V(initial) plug in 0.20 M. For V(final), plug in 4.0 L, and M(final) is unknown. Solve for M, and you get 0.050 M. To find the number moles of NH4NO3 received, multiple 0.050M by 0.1 L. The amount of NH4NO3 is 0.0050 moles. For every mole of NH4NO3, there are two moles of N. Therefore, the plant receives 0.01 moles of N.


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