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how to prepare 0.1n HCL

how to prepare 0.1n HCL

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1 Answers

Aman Bansal
592 Points
11 years ago

Dear Preetha,

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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