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

  • Thread starter Thread starter assaftolko
  • Start date Start date
  • Tags Tags
    Concentration
Click For Summary
SUMMARY

The discussion focuses on calculating the volume of a 32% HCl solution required to prepare a final volume of 300 mL with a molar concentration of 0.1 M. The participants clarify that the density of the stock solution (1.159 g/mL) is relevant for determining the mass of HCl needed. The calculation reveals that 1.095 g of HCl is necessary, which corresponds to a mass of approximately 219 g of the stock solution. This mass is then used to find the volume of the stock solution needed for dilution.

PREREQUISITES
  • Understanding of molarity and solution concentration
  • Knowledge of weight percentage calculations
  • Familiarity with density and its role in solution preparation
  • Basic algebra for solving equations
NEXT STEPS
  • Learn how to calculate molarity from mass and volume
  • Study the principles of dilution and concentration adjustments
  • Explore the weight percentage formula in solution chemistry
  • Investigate the properties of hydrochloric acid solutions
USEFUL FOR

Chemistry students, laboratory technicians, and professionals involved in solution preparation and chemical analysis will benefit from this discussion.

assaftolko
Messages
171
Reaction score
0
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?
 
Physics news on Phys.org


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?
 


assaftolko said:
How am I so far?

Correct.
 


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?
 
  • #10


Yes, yes & yes.
 
  • #11


Borek said:
Yes, yes & yes.

Thank you very much!
 
  • #12
Note: thread had nothing to do with stoichiometry; subject of the first post changed.
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 2 ·
Replies
2
Views
10K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
4
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
2K