Calculate 0.98M CaCl2 to 0.23M Cl- Solution (mL)

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Discussion Overview

The discussion revolves around calculating the volume of 0.98 M calcium chloride needed to prepare 341 mL of a 0.23 M chloride ion solution. It involves stoichiometric relationships and dilution calculations.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant suggests calculating the moles of chloride ions produced by calcium chloride as a first step.
  • Another participant agrees that determining the moles of chloride ions is a necessary part of the process.
  • A participant presents a calculation showing that 0.078 moles of chloride ions are needed and derives a volume of 40.0 mL of calcium chloride based on stoichiometric relationships.
  • A later reply indicates that the calculation appears correct, but does not confirm it definitively.
  • One participant expresses concern about posting in the correct section of the forum.

Areas of Agreement / Disagreement

There is no explicit consensus on the correctness of the final answer, although one participant indicates that the calculation looks okay. The discussion remains somewhat unresolved regarding the final verification of the answer.

Contextual Notes

Participants do not clarify certain assumptions, such as the completeness of the reaction or the conditions under which the calculations are valid.

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Calculate the following quantity:
Volume of 0.98 M calcium chloride (mL) that must be diluted with water to prepare 341 mL of a 0.23 M chloride ion solution.
 
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You have to try first.
 
If I understand the problem correctly, I should find how many mol of chloride ions is produced by one mol of calcium chloride first. DO you think that is right?
 
That's possible approach. For sure you have to do it at some stage.
 
This is how i worked it out:
CaCl2 + 2H2O => Ca(OH)2 + 2HCl

Mol of Cl:
341mL * (1L/1000mL) * 0.23 mol/L= 0.078 mol of Cl

Vol. of CaCl2:
0.078 mol Cl * (1mol CaCl2/2mol Cl) * (1L/0.98 mol CaCl2) * (1000mL/1L) = 40.0 mL CaCl2

I TRIED HARD ON THIS... IS THIS THE RIGHT ANNSWER!
 
Looks OK.
 
Really!? Thanks.

By the way. I think I post this in the wrong section. If you can, please move this post to the proper section. I will read the regulations and follow them the next times.
 

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