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iconwin
Sep21-08, 03:59 PM
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.
iconwin
Sep21-08, 04:14 PM
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.
iconwin
Sep21-08, 05:07 PM
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!
iconwin
Sep21-08, 05:36 PM
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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