document.write( "Question 778238: What quantity of pure acid must be added to 600 mL of a 50% acid solution to produce a 75% acid solution? \n" ); document.write( "

Algebra.Com's Answer #474496 by josgarithmetic(39617) $\"\"$ $\"About$

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Give a variable to how much pure acid.
\n" ); document.write( "Let y = volume in mL of pure acid to add.\r
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\n" ); document.write( "\n" ); document.write( "Note that amount of acid in mixture or resulting solution will be $\"600%2A0.5%2By%2A1\"$ and volume of resulting solution will be $\"y%2B600\"$ mL. We must assume you mean your concentration as volume/volume; otherwise, you will need the densities of your pure acid and of the 50% solution. \r
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\n" ); document.write( "\n" ); document.write( "In your simplified description, your equation is $\"highlight%28%28600%2A0.5%2By%29%2F%28y%2B600%29=0.75%29\"$ ;
\n" ); document.write( " $\"%28300%2By%29=0.75%28y%2B600%29\"$
\n" ); document.write( " $\"y%2B300=0.75y%2B%283%2F4%29600\"$
\n" ); document.write( " $\"y%2B300=0.75y%2B450\"$
\n" ); document.write( " $\"%281%2F4%29y=150\"$
\n" ); document.write( " $\"y=4%2A150\"$
\n" ); document.write( " $\"highlight%28y=600%29\"$ mL of the pure acid. \n" ); document.write( "

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