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Chemistry 12 Online
OpenStudy (anonymous):

Hey super smart chemistry people :D please help me with one tiny little problem? The question is:Calculate the molarity of 1.60 L of a solution containing 1.55g of dissolved KBr. I have been trying to figure this out but I dont know where to start....

OpenStudy (anonymous):

DIvide the mass of KBr by the molecular weight, that's the number of KBr ion-pairs you dissolved, in moles. Divide the number of moles by the volume, that's the molarity, in moles per liter. would you like to try it out here?

OpenStudy (anonymous):

Umm sure?

OpenStudy (anonymous):

@monet212 Try this formula: \[Molarity (M)=\frac{solute (Moles) }{solvent (Liters) }\]

thomaster (thomaster):

@monet212 answered it in your last question: http://openstudy.com/study#/updates/51749a79e4b0050fabb96202

OpenStudy (anonymous):

I know the formula, its just I dont know which is which and what to start with...

OpenStudy (anonymous):

Thanks @thomaster :) I checked my last question and saw your answer

OpenStudy (anonymous):

Oh and thank you @SweetsAndEverything i messed up the formula, you showed me the right one

OpenStudy (anonymous):

No problem. :)

thomaster (thomaster):

Did you manage to work it out? or do you need more help with the intermediate steps

OpenStudy (anonymous):

Umm ya I think I got the hang of it :) identifying each thing is just really confusing me

OpenStudy (anonymous):

Also is there a website that posts videos about this stuff? Because I have 4 more questions that I need help with but I dont want to take up too much of anyones time

thomaster (thomaster):

youtube has a lot of videos explaining this stuff. If you'd like i can search some videos about this. Also the people here like to spend time on helping people :) It's not like you're wasting their time so you can ask all your questions if you need help with them.

thomaster (thomaster):

http://www.youtube.com/watch?v=yb4FW6E1HKE is a good one

OpenStudy (anonymous):

Oh good haha I was starting to feel bad. Thanks for the link and I'm going to ask a few more questions :)

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