Calculating Molar Concentration of Diluted HCl Solution

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To determine the molar concentration of diluted hydrochloric acid after mixing 25 mL of 12 M HCl with 475 mL of distilled water, the formula c1v1 = c2v2 is applied. The initial calculation suggests a concentration of 0.63 M, derived from multiplying 25 mL by 12 M. However, the correct approach considers the total volume of the solution, which is 500 mL. The total moles of HCl in the solution is calculated as 0.3 moles (from 25 mL of 12 M HCl), leading to a final concentration of 0.60 M when divided by the total volume of 0.5 L. This aligns with the textbook answer, confirming that the molarity of the diluted hydrochloric acid is indeed 0.60 M.
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If 25 mL of 12 M HCl are added to 475 mL of distilled water, what is the molar concentration of the diluted hydrochloric acid?

c1v1=c2v2
25*12=475c
c=.63 M HCL

book has the answer as .60 M HCL
Isn't the answer .63?
 
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"25*12=475c"

Think about this a moment; you've asserted something different from the problem statement, "25 mL of 12 M HCl are added to 475 mL ---."
 
Your total volume is 500ml, and you have 12 moles. 12moles/.500L is?
 
this seems to work, .30 mol/.500L = .60 M
 
Originally posted by GeneralChemTutor
Your total volume is 500ml, and you have 12 moles. 12moles/.500L is?

No u have 12/40=0.3 moles
and now volume is 500mL

therefore molarity= 0.3*2=0.60

it is right way fish
 

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