yd3@nvc.net wrote:
> Greetings from cold South Dakota.
>
> A few days ago I conducted an experiment on what happens to weak
> antifreeze mixures when they freeze.
>
> I made up two 500 mL concentrations: 25% antifreeze and 33% antifreeze.
> Now that I am done I wanted to increase each one to 50% antifreeze so I
> can use it in my leaky truck. How much antifreeze should I have added
> to each one to get each mixure back up to 50%?
Case 1
25% antifreeze means the total 500 mL is divided into 125 mL of
antifreeze and 375 mL of water. To get a 50% solution, you need to get
the antifreeze total up to 375 mL so it matches the present water
content. You already have 125 mL in the solution, so you need to add 250
mL more antifreeze.
Case 2
33% antifreeze means the total 500 mL is divided into 166.67 mL of
antifreeze and 333.33 mL of water. To get a 50% solution, you need to
get the antifreeze total up to 333.33 mL so it matches the present water
content. You already have 166.67 mL in the solution, so you need to add
166.67 mL more antifreeze.
Figures rounded off.
-Rock http://www.rocky-frisco.com
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