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?

The answer is:  " 0.420 m " .

Note:  Use the formula:

   m₁V₁ = m₂V₂   ;

   (12.0 m)(0.0350 L) = (m₂)*(12.0 L)  ; solve for "m₂".

    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).

    We have:
     
     (12.0 m)(0.0350 L)  =  (m₂)*(12.0 L)  ; solve for "m₂".

     ↔  (m₂)*(12.0 L)  =  (12.0 m)(0.0350 L) ;

                 →  Divide EACH SIDE of the equation by "(12.0 L)" ;
   to isolate "(m₂)" on one side of the equation; and to solve for  "(m₂)" ;

       → [ (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 ;

                 →   The answer is:  " 0.420 m " .