Question

If i start with 0.0350 l of a 12.0m hydrochloric acid solution, and dilute it with water to a new volume of 12.0 l, what is the new concentration?

Answer 1

The answer is:  " 0.420 m " .
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Note:  Use the formula:
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   m₁V₁ = m₂V₂   ;
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   (12.0 m)(0.0350 L) = (m₂)*(12.0 L)  ; solve for "m₂".
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    Since the units for "m" (concentration) are the same;
       (i.e., "molality";  that is:  "moles of solute / kg solvent) ;
       and since units for "V" (volume) are the same; (i.e. "L" , or "Liters");         
       we do not have to convert).
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    We have:
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     (12.0 m)(0.0350 L)  =  (m₂)*(12.0 L)  ; solve for "m₂".
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     ↔  (m₂)*(12.0 L)  =  (12.0 m)(0.0350 L) ;
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                 →  Divide EACH SIDE of the equation by "(12.0 L)" ;
   to isolate "(m₂)" on one side of the equation; and to solve for  "(m₂)" ;
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       → [ (m₂)*(12.0 L) ] / (12.0 L)  =  [ (12.0 m)(0.0350 L) ] / (12.0 L) ;

                                →   m₂ = [ (12.0) (0.0350) ] m  ;

                                →   m₂ = 0.420 m ;
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                 →   The answer is:  " 0.420 m " .
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