Calculate Volume for 0.1M HCl Solution in 300ml at 32%

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What is the volume you need to take from 32% HCl solution in order to get a solution in final volume of 300 ml and with Molar concentration of 0.1 M. The solution density is 1.159 gr/ml.

I don't understand why will we have change... the mass will stay the same won't it? and the density is constant... so why would the volume change?
 
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I guess question is about solution preparation - how much concentrated solution has to be diluted to 300 mL to prepare 0.1M solution. And the density given is that of the stock solution, not of the final one.
 


Borek said:
I guess question is about solution preparation - how much concentrated solution has to be diluted to 300 mL to prepare 0.1M solution. And the density given is that of the stock solution, not of the final one.

Yes I think I understood the question wrong: They want to know how much volume do I need to remove from a 32% HCl solution in order to get a solution with final volume of 300 ml and with conentration of 0.1M. I still don't know how to do this... but shouldn't the density of the solution be constant even if I extract some of the solution from the vessel it's in?
 


As I explained before - IMHO you are not asked about REMOVING part of the solution, but about calculating how much of the solution needs to be taken and DILUTED. You are not interested in what is left, but in what is prepared. Yes, density of what is left is not changing, but it is irrelevant to the question.
 


Borek said:
As I explained before - IMHO you are not asked about REMOVING part of the solution, but about calculating how much of the solution needs to be taken and DILUTED. You are not interested in what is left, but in what is prepared. Yes, density of what is left is not changing, but it is irrelevant to the question.

you could be right, it's hard to understand their intent from how they wrote the question (which is not in English). Assuming you are correct - can you please help me to understand what are the relevant quantaties I'm interested with in order to solve this prob?
 


How much HCl will be present in the solution after the dilution?

How much of the stock solution contains this amount of HCl?
 


Borek said:
How much HCl will be present in the solution after the dilution?

How much of the stock solution contains this amount of HCl?

If the concentration is 0.1M and the volume of the final solution is 300 ml - then we get:

0.1 * 0.3 = 0.03 mol of HCl in the final solution. Since the molacular weight of HCl is about 36.5 gr/mol, the mass of 0.03 mol of HCl in the final solution is: 0.03 * 36.5 = 1.095 gr.

The stock solution (I hope I understood this defenition) contains 32% HCl, so according to the weight precentage formula, in order to get 1.095 gr of HCl we have:

32 = 1.095/m * 100 where m is the mass of the stock solution. After some algebra:

m= 219/64 gr.

How am I so far?
 


Borek said:
Correct.

So in order to continue I need to divide this mass in the given solution density, so the density is relevant to this question is it not?

basically what we did here was to take some amount of HCl 32% solution, which had the given density, and to dilute it so the final solution had concetration of 0.1M and volume of 300 ml?
 
Note: thread had nothing to do with stoichiometry; subject of the first post changed.