somebody Posted October 24, 2005 Report Share Posted October 24, 2005 Alright, before you read the problem just wanted to make things clear that the problem that i am goona give it to you is a advance REVIEW(advance for me atleast) problem in my book. SO dont think that i am trying to get my homework done here (had some problems in the past with the homework stuff so just want to make clear), its just a review problem before my midterm so it would be nice if you guys give me some good hints :) or steps!! what volume of 0.248 M CaCl2 must be added to 335 ml of 0.186 M KCl to produce a solution with a concentration of 0.250 M Cl-? Easy for you but hard for me! :) Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted October 24, 2005 Report Share Posted October 24, 2005 ok, well if you look at the amount of mol and the volumes seperately initially 0.335L and 0.0623molyour other solution for every L has 0.496mol and it has to equal 0.250M now you could set up an equation 0.25 = n/V 0.25 = (0.0623 + 0.496V)/(0.335 + V) where V = volume added following through to conclusion V = 87.2mL I skipped a few steps setting up the equation and then solving it so if you dont understand just ask :) Tormod 1 Quote Link to comment Share on other sites More sharing options...
somebody Posted October 26, 2005 Author Report Share Posted October 26, 2005 Oh i never tried attepting to solve this problem that way! Thanks man (or woman), it really gave me a hard time. Thanks, I apreciate your help!! Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted October 27, 2005 Report Share Posted October 27, 2005 anytime :) (and its man) Quote Link to comment Share on other sites More sharing options...
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