Understanding question : Concentration

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Discussion Overview

The discussion revolves around a homework problem involving the dilution of a solution. Participants explore the calculations needed to determine the final concentration after a series of dilutions, addressing the assumptions related to volume and concentration changes during the process.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant notes the initial conditions of the problem, stating that 63.0 mL of a 1.70 M solution is diluted to a total volume of 248 mL.
  • Another participant emphasizes the importance of considering the new concentration after the first dilution and questions the interpretation of the water added in the second dilution.
  • A third participant mentions that dilutions are based on mass conservation, implying that the amount of solute remains constant throughout the process.
  • Some participants suggest using the equation ##M_1V_1=M_2V_2## to express the relationship between the concentrations and volumes before and after dilution.

Areas of Agreement / Disagreement

Participants express differing interpretations of the problem, particularly regarding the understanding of volume contributions and concentration changes. There is no consensus on the correct approach to the calculations involved.

Contextual Notes

Participants highlight potential confusion regarding the additive nature of volumes and the implications for concentration calculations, indicating that assumptions about volume and concentration may need clarification.

brycenrg
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Homework Statement


63.0ml of a 1.70 M solution is diluted to a total volume of 248mL. A 124-mL portion of that solution is diluted by adding 167mL of water. What is the final concentration? Assume the volumes are additive

Homework Equations


What is going on in this equation?

The Attempt at a Solution


You have 248 mL solution but 63mL of that solution is with the molarity of 1.7.
When it says 124mL portion of that solution is diluted by adding 167mL of water. That would mean of the 248 mL, 167mL is water but it initially says 63mL is a solution. so that would leave extra water. I must be thinking of this wrong. any suggestions?
 
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brycenrg said:
You have 248 mL solution but 63mL of that solution is with the molarity of 1.7.
That's a very strange way to think about it. You have to consider the molarity of the solution after dilution.

brycenrg said:
When it says 124mL portion of that solution is diluted by adding 167mL of water. That would mean of the 248 mL, 167mL is water but it initially says 63mL is a solution. so that would leave extra water.
I donät get what you are saying. You start with 63 mL of a solution, then dilute that to 248 mL. That gives you 248 mL of solution with a new concentration.

You then take 124 mL of that solution, and dilute it by adding 167 mL of water. That gives you x mL of a solution with a new concentration.

You can basically see the second part of the question as the same as the first part, except the numbers have changed.
 
Dilutions are just about mass conservation - whatever you put in, is present in the final solution.
 
brycenrg said:

Homework Statement


63.0ml of a 1.70 M solution is diluted to a total volume of 248mL. A 124-mL portion of that solution is diluted by adding 167mL of water. What is the final concentration? Assume the volumes are additive.
You start with a big jug of 1.70 M solution. You draw out 63.0 ml of this solution into a beaker.

You take this beaker with the 63.0 mL of 1.70 M solution and dilute it so that the total volume is now 248 mL.

Take half of this diluted solution, 124 mL, and pour it into a second beaker. Add 167 mL water to this portion of already diluted solution, diluting it even more. What's the final concentration?

Even in chemistry HW, sometimes it helps if you make a sketch of what's going on.
 
I suppose the ##M_1V_1=M_2V_2## equation would come in handy, here?
 
Eclair_de_XII said:
I suppose the ##M_1V_1=M_2V_2## equation would come in handy, here?

Yes, that's just a way of expressing mass conservation in the simplest case.
 

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