Understanding question : Concentration

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The discussion revolves around calculating the final concentration after a series of dilutions. Initially, 63.0 mL of a 1.70 M solution is diluted to 248 mL, resulting in a new concentration. A 124 mL portion of this diluted solution is then further diluted by adding 167 mL of water. Participants emphasize the importance of understanding mass conservation in dilutions and suggest using the equation M1V1 = M2V2 to find the final concentration. Clarifications are made regarding the volumes involved and the steps to achieve the correct concentration after each dilution.
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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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