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Chemistry, mL required to Concentration of HNO3

  • Thread starter Flip
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  • #1
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Homework Statement



How many mL of 50.0% by mass HNO3 solution, with a density of 2.00 g/mL is required to make a 500 mL of a 2.00 M HNO3?


Homework Equations



initial Molarity X initial Volume = final Molarity X final Volume


The Attempt at a Solution



(50 g HNO3)(1 mL/2.00 g)(0.001 L/1 mL) = 0.025 L
(50 g HNO3)(1 mol HNO3/63.012 g HNO3) = 0.79 mol

0.79 mol/0.025 L = 31.6 M

(500 mL)(2 M) = (31.6 M)V

V = 31.6 mL


Comments

I don't know if the process that I used is right. Which leads me to doubt the answer. Any help is much appreciated.
 

Answers and Replies

  • #2
Borek
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Generally not bad, although you did one serious mistake:

(50 g HNO3)(1 mL/2.00 g)(0.001 L/1 mL) = 0.025 L
That would be true if you would start with pure acid, not with the solution. 50 g of the solution has volume of 0.025L, that's OK, but it doesn't contain 50 g of HNO3.
 
  • #3
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Thanks for the reply.

Ok so the volume I obtained is right, but the label is wrong? And the answer is around 31.6 mL?
 
  • #4
Borek
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28,441
2,837
No, this is a serious mistake (serious in terms of the way it changes final result).
 
  • #5
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Ok ok, so since it says 50 percent by mass HNO3 I assumed that it was part of a 100 percent solution. Meaning 100 grams. So I did the whole process again with the new calculations and got 63.0 mL.

Here is my work:

(50 g solution)(1 mL/2.00 g)(0.001 L/1 mL) = 0.05 L
(50 g HNO3)(1 mol HNO3/63.012 g HNO3) = 0.79 mol

0.79 mol/0.05 L = 15.8 M

(500 mL)(2 M) = (15.8 M)V

V = 63.0 mL

Which I found out later, from the answer key that my professor provided me with, was correct. Thanks!
 

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