Post Assessment Question

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Yadira Flores 1G
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Joined: Wed Nov 15, 2017 3:01 am

Post Assessment Question

Postby Yadira Flores 1G » Tue Apr 10, 2018 3:21 pm

A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution? Why isn't it correct to apply the M1V1=M2V2 to this specific problem?

BrendaC_1F
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Joined: Wed Nov 16, 2016 3:04 am

Re: Post Assessment Question

Postby BrendaC_1F » Wed Apr 11, 2018 1:05 am

I still did the MiVi=MfVf for this problem. But before you can do that you need to turn the given grams into MOLARITY. Which you can do by first turning the grams(55.1gKCl) into moles. You should get 0.739 mol KCl. Then you divide the moles by the initial Volume given (0.075L). The value you get should be 9.85 M. Then you could start plugging in everything onto the formula in order to solve for the Mfinal.


I hope this is helpful.

Paywand Baghal
Posts: 32
Joined: Fri Apr 06, 2018 11:01 am

Re: Post Assessment Question

Postby Paywand Baghal » Wed Apr 11, 2018 2:20 pm

BrendaC_1F wrote:I still did the MiVi=MfVf for this problem. But before you can do that you need to turn the given grams into MOLARITY. Which you can do by first turning the grams(55.1gKCl) into moles. You should get 0.739 mol KCl. Then you divide the moles by the initial Volume given (0.075L). The value you get should be 9.85 M. Then you could start plugging in everything onto the formula in order to solve for the Mfinal.


I hope this is helpful.


so you would multiply 0.739 mol to 0.075 L (M1V1) then divide it by 0.125 L (V2) to get the final Molarity right? Because M1V1=M2V2, so M2=M1V1/V2


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