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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