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Author Topic: Preparing 0.1 N HCl Solution  (Read 67829 times)

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jbays1973

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Preparing 0.1 N HCl Solution
« on: October 17, 2004, 09:19:06 AM »

I have a question regarding the preparation of this solution.  We are using 36 - 38 % HCl as our stock for the dilution.  I understand that 8.3 mL are required to make the 1 L required.  What I do not understand is how this quantity is arrived at.  I mean, how was it determined that one needs to add 8.3 mL to a liter of DI water to make 0.1 N HCl?  I would appreciate the assistance with this question.  

jim
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AWK

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Re:Preparing 0.1 N HCl Solution
« Reply #1 on: October 17, 2004, 07:32:39 PM »

38 % HCL shows density 1.19 g/mL. Hence 1000 mL weights 1190 grams and contains 0.38 x 1190 = 452.2 g of HCl (12.39 moles per 1L).
Using equation M1V1 = M2V2 you can obtain
V1=0.1 x 1000 /12.39 = 8,1 mL (for 36 % HCl  V1= 8.6 mL)
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