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

[iconwin]

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.

 

[Borek]

You have to try first.

 

[iconwin]

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?

 

[Borek]

That's possible approach. For sure you have to do it at some stage.

 

[iconwin]

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!

 

[Borek]

Looks OK.

 

[iconwin]

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.