My exam is tomorrow and so I started to do some review and I have no idea how to solve these 2 problems.
Determine the volume of concentrated hydrochloric acid (11.6M) required to prepare 2.0L of a solution at 1.0 Mol/L.
What volume of concentrated ammonia is needed to prepare 2.0L of solution at 1.0 Mol/L
The answers for the first one is .172L and for the 2nd one it's .14L. I'm just not sure how to get those answers.
Determine the volume of concentrated hydrochloric acid (11.6M) required to prepare 2.0L of a solution at 1.0 Mol/L.
What volume of concentrated ammonia is needed to prepare 2.0L of solution at 1.0 Mol/L
The answers for the first one is .172L and for the 2nd one it's .14L. I'm just not sure how to get those answers.
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The first question is worked out like so:
2.0L HCl x 1mol HCl/1L HCl x 1L HCl/11.6mol HCl
When you multiple and divide through, you get an answer of 0.172 L
So it's the total volume, multiplied by the end molarity, divided by the starting molarity. If you make sure to cross out all your units until you end up with what you are looking for, the stoichiometry shouldn't be too hard.
The second problem is done the exact same way. All you need to know is the concentration of "concentrated ammonia". This is around 14.8M. If you use that number, you will get 0.135, which can be rounded to 0.14L, the answer you were given.
Good luck with your exam!
2.0L HCl x 1mol HCl/1L HCl x 1L HCl/11.6mol HCl
When you multiple and divide through, you get an answer of 0.172 L
So it's the total volume, multiplied by the end molarity, divided by the starting molarity. If you make sure to cross out all your units until you end up with what you are looking for, the stoichiometry shouldn't be too hard.
The second problem is done the exact same way. All you need to know is the concentration of "concentrated ammonia". This is around 14.8M. If you use that number, you will get 0.135, which can be rounded to 0.14L, the answer you were given.
Good luck with your exam!
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The formula to dilute a solution of anything is:
M(c)V(c) = M(d) V(d) where the c stands for concentrated and the d stands for diluted. The M stands for the molarity of the solution and the V stands for the Volume of the solution.
For your first question you will plug in numbers as follows :
(11.6M) (Vc) = (1.0M) (2.0L) . Divide the 2 by 11.6 and you will get 0.172.
M(c)V(c) = M(d) V(d) where the c stands for concentrated and the d stands for diluted. The M stands for the molarity of the solution and the V stands for the Volume of the solution.
For your first question you will plug in numbers as follows :
(11.6M) (Vc) = (1.0M) (2.0L) . Divide the 2 by 11.6 and you will get 0.172.
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I'll do the first one.
If you need 2 L of 1M solution, then you basically will have 2 moles of pure HCl (1M=xmole/2L)
Thus, use a proportion and solve:
11.6moles/1L=2moles/xL
x=.172L
If you need 2 L of 1M solution, then you basically will have 2 moles of pure HCl (1M=xmole/2L)
Thus, use a proportion and solve:
11.6moles/1L=2moles/xL
x=.172L
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1. M1 x V1 = M2 x V2
11.6M x V1 = 1.0M x 2.0L
V1 = 2.0/11.6
11.6M x V1 = 1.0M x 2.0L
V1 = 2.0/11.6