document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=508286>Question 508286</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If you need 425mL of a .90% saline solution but only have 25.0% saline, how would you go about making the .90% solution? Do calculation and explain in words. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #340920 by </B><A HREF=/tutors/aboutme.mpl?userid=oberobic><B>oberobic(2304)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oberobic><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=340920\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> With mixture problems you have to determine how much 'pure' stuff you need.<BR>\n" );
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document.write( "You need 425 mL of a 0.90% saline:  So you will have 3.825 mL of 100% saline + 421.175 mL of water.<BR>\n" );
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document.write( "You have only 25% saline solution. That is more than 25 times the strength you need.<BR>\n" );
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document.write( ".25x = 3.825<BR>\n" );
document.write( "x = 3.825/.25<BR>\n" );
document.write( "x = 15.3 mL<BR>\n" );
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document.write( "In 15.3 mL of 25% saline you have all the 'pure' saline you need.<BR>\n" );
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document.write( "Add 425-15.3 = 409.7 mL of water.<BR>\n" );
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document.write( "409.7 mL of water + 15.3 mL of 25% saline = 425 mL of 0.90% saline<BR>\n" );
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document.write( "Done. </SPAN>\n" );
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